Motion - Class 9 Science - Chapter 7 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Motion | NCERT | Science | Class 9
The velocity-time graphs of a car and a scooter are shown in the figure. The difference between the distance travelled by the car and the scooter in 15 s and the time at which the car will catch up with the scooter are, respectively.
A. 112.5 m and 22.5 s B. 337.5 m and 25 s C. 112.5 m and 15 s D. 225.5 m and 10 s
To solve the problem, we need to determine two things: the difference in the distance traveled by the car and the scooter in 15 seconds and the time at which the car will catch up with the scooter.
Step 1: Difference in Distance Traveled in 15 Seconds
Distance traveled by the scooter:
The scooter travels at a constant speed of 30 meters per second.
Distance = Speed × Time = $30 , \text{m/s} \times 15 , \text{s} = 450 , \text{m}$
Distance traveled by the car:
The velocity-time graph of the car is a triangle for the first 15 seconds.
The area under the triangle represents the distance.
For a right triangle, the area is $\frac{1}{2} \times \text{base} \times \text{height}$.
Base = 15 seconds, Height = 45 meters per second.
Distance = $\frac{1}{2} \times 15 , \text{s} \times 45 , \text{m/s} = 337.5 , \text{m}$
Difference in distance:
The difference between the distances traveled by the car and scooter in 15 seconds is:
$\text{Difference} = 450 , \text{m} - 337.5 , \text{m} = 112.5 , \text{m}$
Step 2: Time at Which the Car Will Catch Up with the Scooter
Initial distance difference:
From above, we know this is 112.5 meters.
Equalizing the distances for both vehicles:
The scooter continues to travel at a constant speed of 30 meters per second.
The car, after 15 seconds, travels at a constant speed of 45 meters per second.
The additional time needed for the car to catch up the scooter is the time when their respective distances are equal.
Let's denote this additional time as $ t_0 $:
The displacement of the scooter for $ t_0 $ seconds is:
$30 t_0$ meters
The displacement of the car for $ t_0 $ seconds:
Given that the car has already traveled $337.5$ meters in the first 15 seconds, $337.5 + 45(t_0 - 15)$.
Setting these two distances equal gives us: $$ 30 t_0 = 337.5 + 45(t_0 - 15) $$ $$ 30 t_0 = 337.5 + 45 t_0 - 675 $$ $$ 30 t_0 - 45 t_0 = 337.5 - 675 $$ $$ -15 t_0 = -337.5 $$ $$ t_0 = 22.5 , \text{s} $$
Conclusion
The difference in the distance traveled by the car and scooter in 15 seconds is 112.5 meters.
The time at which the car will catch up with the scooter is 22.5 seconds.
Thus, the correct answer is:
Answer: A. 112.5 m and 22.5 s
The graph between the displacement $x$ and time $t$ for a particle moving in a straight line is shown in the figure. During the intervals $OA$, $AB$, $BC$, and $CD$, the acceleration of the particle is:
OA | AB | BC | CD | |
---|---|---|---|---|
(1) | + | 0 | + | + |
(2) | - | 0 | + | 0 |
(3) | + | 0 | - | + |
(4) | - | 0 | - | 0 |
To solve this problem, let's analyze the given graph and determine the acceleration during the intervals $OA$, $AB$, $BC$, and $CD$. We'll use the properties of velocity and acceleration in relation to displacement and time.
Basics:
Acceleration $(a)$ is given by the change in velocity $(v)$ over time: $$ a = \frac{dv}{dt} $$
Velocity $(v)$ is given by the change in displacement $(x)$ over time: $$ v = \frac{dx}{dt} $$
Analysis of the graph:
Interval $OA$:
The graph between $O$ and $A$ is increasing.
Since the slope (which represents velocity) is increasing, the velocity itself is increasing.
Therefore, the acceleration is positive in this interval.
Interval $AB$:
The graph between $A$ and $B$ is a horizontal line.
The slope is zero, which means the velocity is constant.
Therefore, the acceleration is zero in this interval.
Interval $BC$:
The graph between $B$ and $C$ is increasing again.
Since the slope is increasing again, the velocity increases.
Therefore, the acceleration is positive in this interval.
Interval $CD$:
The graph between $C$ and $D$ is also increasing but looks steeper than previous intervals.
The slope (velocity) is increasing, meaning the velocity further increases.
Therefore, acceleration is positive in this interval.
Conclusion:
From the analysis, the pattern of the acceleration through each interval is:
$OA$: Positive
$AB$: Zero
$BC$: Positive
$CD$: Positive
Therefore, the correct answer is: Option 1: +0+
Thus, the final answer is A.
A wheel of circumference $C$ is at rest on the ground. When the wheel rolls forward through half a revolution, then the displacement of the point of contact will be:
A. $C \sqrt{\frac{1}{\pi^{2}} + \frac{1}{4}}$
B. $\frac{C}{2}$
C. $\pi \sqrt{C^{2} + 4}$
D. $C \sqrt{\frac{1}{\pi} + \frac{1}{2}}$
To find the displacement of the point of contact when the wheel rolls forward through half a revolution, let’s break down the problem step by step.
Initial Setup:
A wheel of circumference $ C $ is initially at rest on the ground.
The wheel rolls forward through half a revolution.
Understanding the Movement:
When the wheel undergoes half a revolution, the point of contact initially on the ground will move to a position directly above the ground at the top of the wheel.
Calculating the Distances:
The circumference of the wheel is given by $ C $. Hence, the radius $ r $ of the wheel can be expressed as: $$ C = 2\pi r \quad \Rightarrow \quad r = \frac{C}{2\pi} $$
In half a revolution, the wheel covers a distance of $ \frac{C}{2} $ horizontally on the ground (considering half of its circumference).
The vertical distance moved by the point of contact is equal to the diameter of the wheel, which is $ 2r = \frac{C}{\pi} $.
Using the Pythagorean Theorem:
We need to calculate the displacement of the point of contact, which forms the hypotenuse of a right triangle where:
One leg is the horizontal distance $ \frac{C}{2} $
The other leg is the vertical distance $ \frac{C}{\pi} $
Using the Pythagorean theorem, the displacement (s) is given by: $$ s = \sqrt{\left( \frac{C}{2} \right)^2 + \left( \frac{C}{\pi} \right)^2} $$
Simplifying the Expression:
Simplify inside the square root: $$ s = \sqrt{\frac{C^2}{4} + \frac{C^2}{\pi^2}} $$
Factor out $ C^2 $: $$ s = C \cdot \sqrt{\frac{1}{4} + \frac{1}{\pi^2}} $$
Final Result:
Hence, the displacement of the point of contact when the wheel rolls forward through half a revolution is: $$ s = C \sqrt{\frac{1}{4} + \frac{1}{\pi^2}} $$
Comparing with the given options, the correct answer is in line with:
Option A: $ C \sqrt{\frac{1}{\pi^{2}} + \frac{1}{4}}$
Thus, the correct answer is:
Option A
If $\vec{v}_{m \omega}$ = velocity of a man relative to water, $\vec{v}_{\omega}$ = velocity of water, and $\vec{v}_{m}$ = velocity of man relative to ground, match the following.
Column I | Column II |
---|---|
i. Minimum distance for $v_{m \omega}>v_{\omega}$ | a. $\theta = \sin ^{-1} \left( \frac{\mathrm{V}_{\mathrm{m} u}}{\mathrm{V}_{\omega}} \right)$ |
ii. Minimum time for $v_{m \omega} \geq v_{\omega}$ | b. $\overrightarrow{\mathrm{v}}_{\mathrm{m}} \perp \overrightarrow{\mathrm{v}}_{\omega}$ |
iii. Minimum distance for $v_{m \omega}<v_{\omega}$ | c. $\overrightarrow{\mathrm{v}_{\mathrm{m}} \omega} \perp \overrightarrow{\mathrm{v}}_{\omega}$ |
iv. Minimum time for $v_{m \omega}<v_{\omega}$ | d. $\theta = \sin ^{-1} \frac{v_{\omega}}{v_{m \omega}}$ |
A. i - b, d ii - c iii - a iv - c
B. i - b, d ii - a iii - a iv - c
C. i - b, d ii - c iii - a iv - a
D. i - b, c ii - c iii - a iv - c
The correct option is A: i- b, d; ii- c; iii- a; iv- c.
To determine this, let's analyze each condition in the given problem:
For Column i: Minimum distance for $v_{m \omega} > v_{\omega}$
The minimum distance occurs when the man travels directly from point A to point B. The conditions that align with minimizing distance are: $$ \sin \theta = \frac{v_{\omega}}{v_{m}} $$ and $$ \vec{v}_{m} \perp \vec{v}_{\omega} $$ Thus, i corresponds to b and d.
For Column ii: Minimum time for $v_{m \omega} \geq v_{\omega}$
The minimum time is achieved when the velocity of the man with respect to water is perpendicular to the velocity of the water: $$ \overrightarrow{\mathrm{v}_{\mathrm{m}} \omega} \perp \overrightarrow{\mathrm{v}}_{\omega} $$ So, ii corresponds to c.
For Column iii: Minimum distance for $v_{m \omega} < v_{\omega}$
When the velocity of the man with respect to the water is less than the velocity of the water, the shortest distance is described by: $$ \sin \theta = \frac{\mathrm{v}_{\mathrm{m} \omega}}{\mathrm{v}_{\omega}} $$ Hence, iii corresponds to a.
For Column iv: Minimum time for $v_{m \omega} < v_{\omega}$
Similar to column ii, the minimum time is achieved when: $$ \overrightarrow{\mathrm{v}_{\mathrm{m}} \omega} \perp \overrightarrow{\mathrm{v}}_{\omega} $$ Therefore, iv corresponds to c.
Thus, the correct matching is:
A: i- b, d; ii- c; iii- a; iv- c.
As shown in the figure, a wheel ( $r=1 \mathrm{ m}$ ) is performing combined translational and rotational motion with angular velocity $2 \mathrm{ rad/s}$ and the velocity of its centre as $\vec{V}_{\circ}=4 \hat{i} \mathrm{ m/s}$. Find the velocity of point $B$ in $\mathrm{ m/s}$ on the wheel.
A $\overrightarrow{\mathrm{V}}_{\mathrm{B}}=4 \hat{i}+2 \hat{j}$
B $\vec{V}_{B}=2 \hat{i}-4 \hat{j}$
C $\overrightarrow{V_{B}}=4 \hat{i}-2 \hat{j}$
D $\overrightarrow{\mathrm{V}}_{\mathrm{B}}=2 \hat{i}+4 \hat{j}$
The correct option is C: $\vec{V}_{B}=4 \hat{i}-2 \hat{j}$.
The given data includes:
Radius of the wheel: $r = 1 , \text{m}$
Angular velocity of the wheel: $\omega = 2 , \text{rad/s}$
Velocity of the center of the wheel ($\vec{V}_{\circ}$): $4 \hat{i} , \text{m/s}$
To find the velocity of point ( B ) on the wheel, use the formula: $$ \vec{V}_{B} = \vec{V}_{\circ} + \vec{V}_{\text{BO}} $$
Step-by-Step Calculation:
Angular Velocity Vector: According to the right-hand thumb rule: $$ \vec{\omega} = -2 \hat{k} , \text{rad/s} $$
Velocity of Center of Mass (COM) of the wheel: $$ \overrightarrow{\mathrm{V}}_{\mathrm{O}} = 4 \hat{\mathrm{i}} , \text{m/s} $$
Position Vector of Point ( B ) Relative to the Center ( O ): $$ \vec{r}_{\mathrm{BO}} = 1 \hat{\mathrm{i}} $$
Calculate the Velocity Contribution from Rotation: $$ \vec{V}_{\text{BO}} = \vec{\omega} \times \vec{r}_{\text{BO}} $$
Substituting the values: $$ \vec{V}_{\text{BO}} = (-2 \hat{k}) \times (1 \hat{i}) $$
Using the cross product: $$ \vec{V}_{\text{BO}} = -2 (\hat{k} \times \hat{i}) = -2 \hat{j} , \text{m/s} $$
Calculate the Total Velocity of Point ( B ): $$ \vec{V}_{B} = \vec{V}_{\circ} + \vec{V}_{\text{BO}} = 4 \hat{i} + (-2 \hat{j}) $$
This simplifies to: $$ \vec{V}_{B} = 4 \hat{i} - 2 \hat{j} , \text{m/s} $$
Thus, the velocity of point $B$ is $ \vec{V}_{B}=4 \hat{i} - 2 \hat{j} , \text{m/s} $.
A stone tied to one end of a string 1 m long is whirled in a horizontal circle with a constant speed. If the stone makes 10 revolutions in 20 s, the magnitude of the acceleration of the stone is close to
(A) $98.6 \frac{\mathrm{cm}}{\mathrm{s}^{2}}$ B $490 \frac{\mathrm{cm}}{\mathrm{s}^{2}}$ C $987 \frac{\mathrm{cm}}{\mathrm{s}^{2}}$ D $996 \frac{\mathrm{cm}}{\mathrm{s}^{2}}$
The correct option is C: $987 ;\frac{\text{cm}}{\text{s}^2}$
Here's a step-by-step breakdown:
Radius of the circle: Given as $1 \text{ m}$, which is equivalent to $100 \text{ cm}$.
Number of revolutions: The stone makes 10 revolutions in $20 \text{ s}$.
Revolutions per second (frequency, $f$): $$ f = \frac{10 \text{ revolutions}}{20 \text{ s}} = 0.5 \text{ revolutions per second}$$
Angular velocity ($\omega$): $$ \omega = 2 \pi f = 2 \pi (0.5) = \pi \text{ rad/s} $$
Centripetal acceleration: The formula for centripetal acceleration is $$ a = \omega^2 r $$
Substituting the values: $$ a = (\pi)^2 \times 100 ;\frac{\text{cm}}{\text{s}^2} $$ $$ a \approx 987 ;\frac{\text{cm}}{\text{s}^2} $$
Hence, the magnitude of the acceleration of the stone is approximately $987 ;\frac{\text{cm}}{\text{s}^2}$, corresponding to option C.
A disc of radius $r$ is performing rotational and translational motion as shown in the figure.
Choose the correct option:
A Velocity of point A with respect to point B will always remain constant. B Speed of point A with respect to point B will always remain constant. C Magnitude of relative velocity of point A with respect to point B is $2 \omega |\vec{r}_{AB}| , \text{m/s}$. D None of these.
The correct choice is B. Speed of point A with respect to point B will always remain constant.
Let's analyze why:
Given that the disc is undergoing both rotational and translational motion, we can use the relationship: $$ \vec{v}_{\text{AB}} = \vec{\omega} \times \vec{r}_{\text{AB}} $$
Here:
$\vec{\omega}$ is the angular velocity vector.
$\vec{r}_{\text{AB}}$ is the position vector from point B to point A.
As the disc moves, $\vec{r}_{\text{AB}}$ will change due to the rotational and translational motion. However, the magnitude of the angular velocity vector $\vec{\omega}$ remains constant. Consequently, the magnitude of the velocity vector $\vec{v}_{\text{AB}}$, represented by: $$ |\vec{v}_{\text{AB}}| = |\vec{\omega} \times \vec{r}_{\text{AB}}| $$ remains constant as long as the angular speed $\omega$ and distance $|\vec{r}_{\text{AB}}|$ remain unchanged during the motion.
Therefore, since the magnitude of the velocity (which is equal to the speed) between points A and B remains constant throughout the motion, the correct option is B.
Find the acceleration of the 500 g block in the figure.
Given: $\mathrm{m}_{1} = 500 \text{ g} = 0.5 \text{ kg}$
$$ \begin{align*} \mathrm{m}_{2} &= 100 \text{ g} = 0.1 \text{ kg} \ \mathrm{m}_{3} &= 50 \text{ g} = 0.05 \text{ kg} \end{align*} $$
Consider the following Free Body Diagrams (FBDs):
FBD for m₁:
For the 500 g block ($m_1$): $$0.5g - T_1 = 0.5a \quad \Rightarrow \quad (1)$$
FBD for m₂:
For the 100 g block ($m_2$) on the inclined plane: $$T_1 - T_2 = 0.1a \quad \Rightarrow \quad (2)$$
FBD for m₃:
For the 50 g block ($m_3$): $$T_2 - 0.05g = 0.05a \quad \Rightarrow \quad (3)$$
Adding the above equations to eliminate ( T_1 ) and ( T_2 ):
$$ \begin{align*} (0.5g - T_1) &+ (T_1 - T_2) + (T_2 - 0.05g) = 0.5a + 0.1a + 0.05a \ 0.5g - 0.05g &= (0.5 + 0.1 + 0.05)a \ 0.45g &= 0.65a \ a &= \frac{0.45g}{0.65} \end{align*} $$
Using ( g = 9.8 , \text{m/s}^2 ):
$$ a = \frac{0.45 \times 9.8}{0.65} $$
$$ a \approx 6.92 \frac{\text{m}}{\text{s}^{2}} $$
Thus, the acceleration of the 500 g block is downward with $ 6.92 , \text{m/s}^2 $.
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Ask Chatterbot AIExtra Questions and Answers - Motion | NCERT | Science | Class 9
A driver of a car travelling at $52 \mathrm{~km} \mathrm{~h}^{-1}$ applies the brakes. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?
The graph below represents the velocity of a car over time, where the car initially travels at $52 , \text{km/h}$ and then applies the brakes, showing a decreasing exponential trend in velocity. The shaded area under the curve indicates the distance travelled by the car during the period.
(b) The part of the graph representing uniform motion of the car is where the speed-time graph is a horizontal line, indicating that the speed is constant during that time period.
A piece of stone is thrown vertically upwards. It reaches the maximum height in 3 seconds. If the acceleration of the stone be $9.8m/s^2$ directed towards the ground, calculate the initial velocity of the stone with which it is thrown upwards.
To find the initial velocity of the stone, which was thrown vertically upwards, reaching its maximum height in 3 seconds, we use the kinematic equation for vertical motion:
$$ v = u + at $$
Where:
- $ v $ is the final velocity
- $ u$ is the initial velocity
- $ a $ is the acceleration (since it's directed downward, it will be negative because the motion and acceleration are in opposite directions)
- $ t $ is the time
At the maximum height, the final velocity ($ v $) of the stone is 0 $ m/s $ because it momentarily stops before beginning to fall back down. Since the stone was thrown upwards and gravity is acting downwards, the acceleration due to gravity ($ g $) is -9.8 $ m/s^2$ (negative, as it opposes the initial throw). The time taken to reach the maximum height is 3 seconds. Substituting these values into the equation, we get:
$$ 0 = u + (-9.8 \times 3) $$
$$ 0 = u - 29.4 $$
So,
$$ u = 29.4 \, m/s $$
Therefore, the initial velocity $ u $ with which the stone was thrown upwards is 29.4 $ m/s $.
A stone is dropped into water from a bridge 44.1 m above the water. Another stone is thrown vertically downward 1 second later. Both strike the water simultaneously. What was the initial speed of the second stone?
(A) 12.25 m/s
(B) 14.75 m/s
(C) 16.23 m/s
(D) 16.15 m/s
To solve the problem, we begin by calculating how long it takes the first stone to reach the water. We know the distance $h = 44.1 , \text{m}$, and assume the initial velocity $u = 0 , \text{m/s}$ (since the stone is dropped). Using the equation of motion:
$$ h = ut + \frac{1}{2} g t^2 $$
Substituting the values into the equation:
$$ 44.1 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 $$
Solving for $t$:
$$ t = \sqrt{\frac{2 \times 44.1}{9.8}} \approx 3 , \text{sec} $$
Thus, the first stone takes approximately 3 seconds to hit the water.
The other condition given is that the second stone is thrown 1 second later but still reaches the water at the same time as the first stone. This means the second stone takes $3 - 1 = 2 , \text{sec}$ to reach the water. We need to find the initial speed $u$ of the second stone using the same equation of motion but solving for $u$ this time.
Given $h = 44.1 , \text{m}$, time $t = 2 , \text{sec}$, and rearranging the equation:
$$ 44.1 = u \cdot 2 + \frac{1}{2} \cdot 9.8 \cdot (2)^2 $$
Calculate the second term: $$ \frac{1}{2} \cdot 9.8 \cdot 4 = 19.6 $$
So, $$ 44.1 - 19.6 = 2u \Rightarrow 24.5 = 2u \Rightarrow u = \frac{24.5}{2} = 12.25 , \text{m/s} $$
Thus, the initial speed of the second stone was 12.25 m/s. The correct answer is (A) 12.25 m/s.
What do you mean by periodic motion? Give two examples.
Definition of Periodic Motion (1 Mark):
Periodic motion is defined as motion that repeats itself at regular intervals of time.Examples (2 Marks):
Pendulum: Consider a pendulum where a stone is tied to one end of a string, and the other end is held. If the stone is pulled to one side and released, it will swing back and forth. This motion, which repeats at regular intervals, is an example of periodic motion.
Branch of a Tree: The to-and-fro movement of a tree branch, caused by wind or other forces, which repeats regularly, also constitutes periodic motion.
Find the scalar and vector from the following: speed, distance, current, time, velocity, pressure, work done, momentum, and acceleration.
Scalar quantities are defined by having only magnitude and no direction. They do not follow the parallelogram law of vector addition. The scalar quantities from the given list are:
Speed: Measures how fast an object is moving, only has magnitude.
Distance: Refers to how much ground an object has covered during its motion, only magnitude.
Current: Despite having a direction, it is treated as scalar because it does not follow vector addition.
Time: A measure that does not involve direction.
Pressure: Force per unit area applied in a direction perpendicular to the surface of an object.
Work done: Calculated as the dot product of force and displacement, resulting in a scalar.
Vector quantities have both magnitude and direction and obey the parallelogram law of vector addition. The vector quantities in the list are:
Velocity: Speed with a specified direction.
Acceleration: Describes the rate of change of velocity, has both magnitude and direction.
Momentum: Product of the mass and velocity of an object, a vector quantity due to the direction of velocity.
In v-t graph shown in fig, the displacement of the body in 5 seconds will be, if it starts from origin.
A) $90$ m
B) $70$ m
C) $50$ m
D) $60$ m
To calculate the displacement from a velocity-time graph, you find the area under the graph between the limits of time.
This graph consists of two sections:
A constant velocity section from $t = 0$ s to $t = 2$ s where the velocity $v = 20$ m/s.
A linearly decreasing velocity section from $t = 2$ s to $t = 5$ s, decreasing from $v = 20$ m/s to $v = 0$ m/s.
Calculating the Displacement:
Section 1 (from $t = 0$ to $t = 2$ s):
Shape: Rectangle
Height: $v = 20$ m/s
Base: $t = 2$ s
Area (Displacement) = $height \times base = 20\, \text{m/s} \times 2\, \text{s} = 40$ meters
Section 2 (from $t = 2$ to $t = 5$ s):
Shape: Right triangle
Base: $3$ s (from $t = 2$ to $t = 5$ s
Height: $20$ m/s (decline from $20$ m/s to $0$ m/s)
Area (Displacement) = $\frac{1}{2} \times base \times height = \frac{1}{2} \times 3\, \text{s} \times 20\, \text{m/s} = 30$ meters
Total Displacement:
$ Total\, Displacement = Displacement_{1} + Displacement_{2} = 40\, \text{meters} + 30\, \text{meters} = 70$ meters
Therefore, the total displacement of the body over 5 seconds is 70 meters.
The average velocity of a body moving with uniform acceleration after traveling a distance of $3.06 \mathrm{~m}$ is $0.34 \mathrm{~m/s}$. If the change in velocity of the body is $0.18 \mathrm{~m/s}$ during this time, its uniform acceleration is?
To find the uniform acceleration, we start by determining the initial and final velocities ($u$ and $v$ respectively). We are given the average velocity ($v_{\text{avg}}$) and the change in velocity during the motion.
The average velocity is given by: $$ v_{\text{avg}} = \frac{v + u}{2} = 0.34 , \text{m/s} $$ From this, we can deduce that: $$ v + u = 2 \times 0.34 = 0.68 , \text{m/s} $$
The change in velocity is provided as: $$ v - u = 0.18 , \text{m/s} $$
We now have two equations:
$ v + u = 0.68 $
$ v - u = 0.18 $
Solving these equations simultaneously, we add them to eliminate $u$: $$ (v + u) + (v - u) = 0.68 + 0.18 \ 2v = 0.86 \ v = \frac{0.86}{2} = 0.43 , \text{m/s} $$ and substituting back for $u$: $$ v + u = 0.68 \ 0.43 + u = 0.68 \ u = 0.68 - 0.43 = 0.25 , \text{m/s} $$
With $u$, $v$ and the distance $s = 3.06 , \text{m}$ known, we use the kinematic equation to find the acceleration $a$: $$ v^2 = u^2 + 2as \ 0.43^2 = 0.25^2 + 2 \times a \times 3.06 $$ Calculating each term: $$ 0.1849 = 0.0625 + 6.12a $$ Solving for $a$: $$ 6.12a = 0.1849 - 0.0625 \ 6.12a = 0.1224 \ a = \frac{0.1224}{6.12} \approx 0.02 , \text{m/s}^2 $$
Hence, the uniform acceleration $a$ of the body is $0.02 , \text{m/s}^2$.
The blocks $B$ and $C$ in the figure have mass $m$ each. The strings $AB$ and $BC$ are light, having tensions $T_{1}$ and $T_{2}$ respectively. The system is in equilibrium with a constant horizontal force $mg$ acting on $C$. Then:
(A) $\tan \theta_{1}= \frac{1}{2}$
(B) $\tan \theta_{2}= 1$
(C) $T_{1}= \sqrt{5} mg$
(D) $T_{2}= \sqrt{2} mg$
The system in equilibrium reveals that:
Horizontal force balance on block $C$ gives: $$ T_2 \sin \theta_2 = mg \quad \text{(i)} $$ Applying horizontal force balance for potential horizontal components reveals that $T_2 \sin \theta_2$ equals the horizontal force $mg$ acting on $C$.
Vertical force balance on block $C$ gives: $$ T_2 \cos \theta_2 = mg \quad \text{(ii)} $$ This implies that the vertical component of the tension $T_2$ also equals $mg$, the gravitational force on $C$.
From equations $\text{(i)}$ and $\text{(ii)}$, we determine $\theta_2$: $$ \tan \theta_2 = \frac{\sin \theta_2}{\cos \theta_2} = \frac{mg}{mg} = 1 $$ Therefore, (B) $\tan \theta_2 = 1$ is correct.
Considering the tensions and angles for block $B$, we equate the vertical forces: $$ T_1 \sin \theta_1 = T_2 \sin \theta_2 $$ Using $\text{(i)}$: $$ T_1 \sin \theta_1 = mg $$ And the horizontal forces must balance: $$ T_1 \cos \theta_1 = T_2 \cos \theta_2 + mg $$ Plugging in $\text{(ii)}$: $$ T_1 \cos \theta_1 = 2mg $$ Solving for $\tan \theta_1$, we get: $$ \tan \theta_1 = \frac{T_1 \sin \theta_1}{T_1 \cos \theta_1} = \frac{mg}{2mg} = \frac{1}{2} $$ Therefore, (A) $\tan \theta_1 = \frac{1}{2}$ is correct.
Calculating the tensions from the above expressions, we find: $$ T_2 = \sqrt{(mg)^2 + (mg)^2} = \sqrt{2}mg $$ Hence, (D) $T_2 = \sqrt{2}mg$ is correct.
Likewise, we calculate $T_1$: $$ T_1 = \sqrt{(mg)^2 + (2mg)^2} = \sqrt{5}mg $$ Thus, (C) $T_1 = \sqrt{5}mg$ is correct.
Summary: All options (A), (B), (C), and (D) are correct, fulfilling respective conditions in the described equilibrium scenario.
Why is acceleration related to velocity and not speed?
Acceleration is related to velocity because both are vector quantities. This means they have both magnitude and direction. On the other hand, speed is a scalar quantity, which only has magnitude and no direction.
For example, consider a car moving from left to right at a speed of $5 , \text{km/h}$. Its speed is $5 , \text{km/h}$ and so is its velocity, $5 , \text{km/h}$, where the positive sign indicates the direction of movement from left to right. However, if the same car returns from right to left at $5 , \text{km/h}$, its speed remains constant at $5 , \text{km/h}$, but its velocity is now $-5 , \text{km/h}$. Here, the negative sign indicates the car is moving in the opposite direction.
Like velocity, acceleration also has a direction component. This attribute explains why acceleration is closely related to velocity, but not with speed. When a car changes direction, even while maintaining the same speed, its velocity changes due to the direction component – this change can result in acceleration. This directional nature of acceleration connects it with velocity rather than with speed.
At time $t=0$, apple 1 is dropped from a bridge onto a roadway beneath the bridge; somewhat later, apple 2 is thrown down from the same height. The figure below shows the vertical position, $y$, of the apples versus $t$, time of the falling, until both apples have hit the roadway. The scaling is set by $t_{s}=2.0 \mathrm{~s}$. With approximately what speed was apple 2 thrown down?
A) $15 \mathrm{~m} / \mathrm{s}$
B) $20 \mathrm{~m} / \mathrm{s}$
C) $25 \mathrm{~m} / \mathrm{s}$
D) $10 \mathrm{~m} / \mathrm{s}$
The correct option is A
$$ 15 , \text{m/s} $$
To determine the initial velocity $u$ with which apple 2 was thrown, we start by calculating the height from which the apples were dropped. Using the equation for the position $s$ under uniform acceleration $a$ (due to gravity here): $$ s = ut + \frac{1}{2}at^2 $$
For apple 1, which was freely dropped (thus $u=0$), the equation simplifies to: $$ s = \frac{1}{2}(-10) \times (2)^2 = -20 , \text{m} $$
This tells us the height of the bridge above the roadway is 20 m.
For apple 2, which reaches the roadway in half the time (i.e., $t = 1 , \text{s}$), we use the displacement equation with the unknown initial velocity $u$: $$ -20 = u \times 1 + \frac{1}{2}(-10) \times (1)^2 $$ Solving for $u$: $$ -20 = u - 5 \ u = -15 , \text{m/s} $$
The negative sign indicates the direction of motion (downward). Hence, apple 2 was thrown down with a velocity of 15 m/s.
A table clock has its minute hand 4 cm long. Find the average velocity of the tip of the minute hand
(a) between 6 AM to 6:30 AM and
(b) between 6 AM to 6:30 PM.
A) $4 × 10^{-3} ~\text{cm/s}, 1.8 × 10^{-4} ~\text{cm/s}$
B) $4.4 × 10^{-5} \text{cm/s}, 1.8 × 10^{-4} \text{cm/s}$
C) $0.44 × 10^{-3} \text{cm/s}, 1.8 × 10^{-4} \text{cm/s}$
D) $4.4 × 10^{-3} \text{cm/s}, 0.18 × 10^{-4} \text{cm/s}$
To find the average velocity of the tip of the minute hand of a table clock that is 4 cm long:
The minute hand begins at the 12 o'clock position at 6 AM and reaches the 6 o'clock position at 6:30 AM or 6:30 PM, describing an angle of 180°. This trajectory corresponds to moving along the diameter of the circular path traced by the minute hand.
Given:
The radius (R) of the circle traced by the minute hand = 4 cm
Displacement is the straight-line distance from the start to the end position, which equals the diameter of the circular path: $$ \text{Displacement} = 2 \times \text{Radius} = 2 \times 4 , \text{cm} = 8 , \text{cm} $$
Average Velocity Calculation:
From 6 AM to 6:30 AM (Half an hour or 1800 seconds):
Time taken = 1800 seconds
Average velocity is given by: $$ V_{\text{av}} = \frac{\text{Displacement}}{\text{Time}} = \frac{8}{1800} = 4.4 \times 10^{-3} , \frac{\text{cm}}{\text{s}} $$
From 6 AM to 6:30 PM (12.5 hours or 45000 seconds):
Time taken = 45000 seconds
Average velocity is given by: $$ V_{\text{av}} = \frac{\text{Displacement}}{\text{Time}} = \frac{8}{45000} = 1.8 \times 10^{-4} , \frac{\text{cm}}{\text{s}} $$
Correct Answer: Option B ($4.4 \times 10^{-3} , \frac{\text{cm}}{\text{s}}$, $1.8 \times 10^{-4} , \frac{\text{cm}}{\text{s}}$).
A train 760 m long crosses a platform 440 m long in 40 seconds. Find the speed of the train in km per hour.
Solution
To find the speed of the train in kilometers per hour, we first calculate the total distance the train covers when crossing the platform. This is the length of the train plus the length of the platform:
$$ \text{Total distance} = 760 \text{ m} + 440 \text{ m} = 1200 \text{ m} $$
The train takes 40 seconds to cover this distance. Hence, the speed of the train in meters per second is calculated by using the formula for speed:
$$ \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{1200 \text{ m}}{40 \text{ sec}} = 30 \text{ m/s} $$
To convert this speed from meters per second to kilometers per hour, we use the conversion factor where 1 m/s equals $ 3.6 $ km/h:
$$ \text{Speed in km/h} = 30 \text{ m/s} \times 3.6 = 108 \text{ km/h} $$
Therefore, the speed of the train is $ \mathbf{108} $ km/h.
What conclusion can you draw about the acceleration of a body from the speed-time graph shown below?
From the speed-time graph provided, we can conclude that the acceleration of the body is not constant. Furthermore, it indicates that the acceleration of the body is non-uniformly increasing throughout the observed period.
A vehicle is accelerating on a straight road. Its velocity at any instant is $30 \mathrm{kmh}^{-1}$. After $2 \mathrm{~s}$, it is $33.6 \mathrm{kmh}^{-1}$, and after a further $2 \mathrm{s}$, it is $37.2 \mathrm{kmh}^{-1}$. Find the acceleration of the vehicle in $\mathrm{ms}^{-2}$. Is the acceleration uniform?
Step 1: Calculate the acceleration for the first time interval
The initial velocity is $30 \mathrm{~km/h}$ and after $2 \mathrm{~s}$, it becomes $33.6 \mathrm{~km/h}$. The change in velocity is therefore:
$$
\Delta v = 33.6 \mathrm{~km/h} - 30 \mathrm{~km/h} = 3.6 \mathrm{~km/h}
$$
To convert this change in velocity to meters per second, multiply by $\frac{5}{18}$:
$$
\Delta v = 3.6 \times \frac{5}{18} \mathrm{~m/s} = 1 \mathrm{~m/s}
$$
Now, calculate the acceleration using the formula $a = \frac{\Delta v}{\Delta t}$:
$$
a = \frac{1 \mathrm{~m/s}}{2 \mathrm{~s}} = 0.5 \mathrm{~m/s}^2
$$
Step 2: Calculate the acceleration for the second time interval
The velocity after 2 more seconds (total of 4 seconds) reaches $37.2 \mathrm{~km/h}$. Again, calculate the change in velocity:
$$
\Delta v = 37.2 \mathrm{~km/h} - 33.6 \mathrm{~km/h} = 3.6 \mathrm{~km/h}
$$
Convert to meters per second:
$$
\Delta v = 3.6 \times \frac{5}{18} \mathrm{~m/s} = 1 \mathrm{~m/s}
$$
Again, calculate the acceleration:
$$
a = \frac{1 \mathrm{~m/s}}{2 \mathrm{~s}} = 0.5 \mathrm{~m/s}^2
$$
Conclusion:The acceleration is consistent at $0.5 \mathrm{~m/s}^2$ over both time intervals, suggesting that the acceleration is uniform.
Water is filled in a uniform container of the area of cross section A. A hole of cross section area a ($<<A$) is made in the container at a height of 20 m above the base. Water streams out and hits a small block placed at some distance from the container. With what speed (in m/s) should the block be moved such that the water stream always hits the block? (Given $\frac{a}{A}=\frac{1}{20}$). (Take $g=10 \text{ m/s}^2$)
To solve this problem, we need to determine the velocity at which a block should move such that a stream of water from a container consistently hits it. We can do this by employing the principles of fluid dynamics and motion.
Step 1: Calculate the Velocity of Water Efflux
The velocity of the water efflux, $v$, as it exits through a hole at a height $y = 20, \text{m}$ from the base can be determined using the Torricelli's formula: $$ v = \sqrt{2gy} $$ where $g = 10 , \text{m/s}^2$.
Step 2: Analyze the Continuity Equation
For a fluid in motion, the continuity equation relates the areas and velocities at different points. For an area $A$ of the container and a much smaller hole area $a$, with $\frac{a}{A} = \frac{1}{20}$, the equation is: $$ A \frac{dy}{dt} = a \sqrt{2gy} $$
Step 3: Derive the Equation of Motion for the Block
When the water exits the hole, it will travel in a projectile motion. To find the velocity of the block such that it matches the horizontal position of the water stream, differentiate the horizontal position $x$ with respect to time $t$: $$ V_b = \frac{dx}{dt} $$ We find $x$ using projectile motion equations: $$ x = v \sqrt{\frac{2h}{g}} = \sqrt{2gy} \times \sqrt{\frac{2h}{g}} $$ Differentiating $x$ with respect to time, we have: $$ V_b = \frac{\sqrt{h}}{\sqrt{y}} \cdot \frac{dy}{dt} $$
Step 4: Plug in the Continuity Equation
Substituting the continuity equation into the expression for $V_b$, we get: $$ V_b = \sqrt{\frac{h}{y}} \times \frac{a}{A} \sqrt{2gy} $$
Step 5: Simplify and Calculate $V_b$
Substituting $\frac{a}{A} = \frac{1}{20}$ and simplifying, the equation for $V_b$ becomes: $$ V_b = \sqrt{2gh} \times \frac{a}{A} = \sqrt{2g \cdot 20, \text{m}} \times \frac{1}{20} $$ With $g = 10 , \text{m/s}^2$: $$ V_b = \sqrt{400} \times \frac{1}{20} = 20 \times \frac{1}{20} = 1 , \text{m/s} $$
Conclusion
The block must move at a constant velocity of $1, \text{m/s}$ so that the stream of water from the container always hits it.
A block of mass $1 \mathrm{~kg}$ is pushed towards another block of mass $2 \mathrm{~kg}$ from $6 \mathrm{~m}$ distance as shown in the figure. Just after the collision, the velocity of the $2 \mathrm{~kg}$ block becomes $4 \mathrm{~m/s}$. The coefficient of restitution between the blocks is:
A) 1
B) 0.75
C) 0.5
D) 0.25
To determine the coefficient of restitution ( e ) between the two blocks, we use the given post-collision velocity of the $2 , \text{kg}$ block and the known initial conditions.
Given:
Mass of the first block, ( m_1 = 1 , \text{kg} )
Initial velocity of the first block, ( u_1 = 6 , \text{m/s} ) (towards the second block)
Mass of the second block, ( m_2 = 2 , \text{kg} )
Initial velocity of the second block, ( u_2 = 0 , \text{m/s} ) (stationary)
Final velocity of the second block post-collision, ( v_2 = 4 , \text{m/s} )
Coefficient of Restitution Formula:The formula to determine the velocity of the second object after the collision, assuming the first object comes to rest, is: $$ v_2 = \frac{(1+e) m_1 u_1}{m_1 + m_2} $$ Plugging in the given values: $$ 4 = \frac{(1+e) \times 1 \times 6}{1 + 2} $$ Simplifying the right side: $$ 4 = \frac{6 + 6e}{3} \ 4 = 2 + 2e \ 2 = 2e \ e = 1 $$
Conclusion:The correct answer is Option A: 1. This means the collision is perfectly elastic, as the coefficient of restitution ( e ) equals 1.
2 plastic balls have masses of $100 \mathrm{~g}$ and $200 \mathrm{~g}$. The $100 \mathrm{~g}$ ball is moving at $3 \mathrm{~ms}^{-1}$ and hits the $200 \mathrm{~g}$ stationary ball. They collide with each other and after the collision, the $100 \mathrm{~g}$ ball moves with a velocity of $1.5 \mathrm{~ms}^{-1}$ in the same direction. Calculate the velocity of the other ball after the collision.
A. $0.75 \mathrm{~ms}^{-1}$
B. $0.75 \mathrm{kms}^{-1}$
C. $0.75 \mathrm{cms}^{-1}$
D. $0.57 \mathrm{~ms}^{-1}$
The correct answer is Option A: $0.75 \mathrm{~ms}^{-1}$.
For the motion scenario involving the two plastic balls, let's define:
Mass of the first ball, $m_1 = 100 \mathrm{~g} = 0.1 \mathrm{~kg}$.
Mass of the second ball, $m_2 = 200 \mathrm{~g} = 0.2 \mathrm{~kg}$.
Initial velocity of the first ball, $u_1 = 3 \mathrm{~ms}^{-1}$.
Initial velocity of the second ball (which is stationary), $u_2 = 0 \mathrm{~ms}^{-1}$.
Final velocity of the first ball after the collision, $v_1 = 1.5 \mathrm{~ms}^{-1}$.
Final velocity of the second ball after the collision needs to be determined as $v_2$.
By the conservation of momentum, the total momentum before and after the collision must be equal, so we have:
$$ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 $$
Plugging in the values:
$$ 0.1 \cdot 3 + 0.2 \cdot 0 = 0.1 \cdot 1.5 + 0.2 \cdot v_2 $$
$$ 0.3 = 0.15 + 0.2 v_2 $$
Rearranging for $v_2$:
$$ 0.2 v_2 = 0.3 - 0.15 $$
$$ v_2 = \frac{0.15}{0.2} = 0.75 \mathrm{~ms}^{-1} $$
Thus, the second ball's velocity after the collision is $0.75 \mathrm{~ms}^{-1}$.
$$ \begin{array}{l} V = 3 + 9t \text{ if } t < 9 \ V = 8 - 2t \text{ if } t = 10 \end{array} $$
Find the value of $V(\text{m/s})$ when $t = 10$.
A. $V = 10$ m/s
B. $V = -10$ m/s
C. $V = -5$ m/s
D. $V = -12$ m/s
The correct calculation of $V$ when $t = 10$ s is required, as specified in the given piecewise functions:
$V = 3 + 9t \text{ if } t < 9$
$V = 8 - 2t \text{ if } t = 10$
Since we are asked to find $V$ at $t = 10$, we use the second equation:
$$ V = 8 - 2(10) $$
Calculating this,
$$ V = 8 - 20 = -12 \text{ m/s} $$
Thus, the value of $V$ when $t = 10$ s is $V = -12$ m/s, which corresponds to option D.
Choose the correct statement(s):
A. The average velocity of a moving body can be zero. B. The average velocity of a moving body cannot be zero. C. The average speed of a moving body can be zero. D. The average speed of a moving body is always zero.
Solution
The correct statements are:
A: The average velocity of a moving body can be zero.
Average velocity is calculated using the formula: $$ \text{Average velocity} = \frac{\text{Displacement}}{\text{Time}} $$ The average velocity of a moving body can indeed be zero. This occurs, for example, when the body returns to its starting position after some duration, making the displacement zero. Thus, regardless of the time taken, if the displacement is zero, the average velocity is also zero.
Statement D is incorrect, and the proper interpretation should be:
The average speed of a moving body cannot be zero.
Average speed is defined as: $$ \text{Average speed} = \frac{\text{Distance covered}}{\text{Time}} $$ Since a moving body covers some distance, the average speed must be greater than zero. Therefore, it is impossible for the average speed of a moving body to be zero as it implies no distance was covered, which contradicts the definition of moving.
"Are motion and rest relative terms? Please give one example."
Rest and motion are indeed relative terms, dependent on the frame of reference chosen. Without a point of reference, one cannot precisely state whether an object is at rest or in motion.
Consider the following example:
Imagine you are traveling on a bus. To you, all the other passengers appear to be at rest because they are not moving relative to you. However, to an observer standing on the roadside, everyone on the bus, including you, is moving.
In this scenario:
From your perspective (inside the bus), the passengers are at rest.
From the external observer's perspective (outside the bus), the same passengers are in motion.
This example clearly shows how motion and rest are relative concepts that depend on the observer’s frame of reference.
A driver is driving a car at a speed of $72 \mathrm{~km/hr}$ on the road. He sees a boy on the road at a distance of $48 \mathrm{~m}$ from him. If his reaction time is $0.4 \mathrm{sec}$, find the minimum retardation required to save the boy.
A) $4 \mathrm{~m/s}^{2}$
B) $4.4 \mathrm{~m/s}^{2}$
C) $3 \mathrm{~m/s}^{2}$
D) $5 \mathrm{~m/s}^{2}$
The correct answer is: D) $5 \mathrm{~m/s}^{2}$
When the driver reacts, he travels an additional distance during the reaction time. Given the car's speed at $72 \mathrm{~km/hr}$, we first convert this speed into meters per second: $$ \text{Speed in (\mathrm{m/s})} = 72 \times \frac{1000 , \mathrm{m}}{3600 , \mathrm{s}} = 20 , \mathrm{m/s} $$
Given his reaction time is $0.4 \mathrm{~sec}$, the distance covered in this time is: $$ \text{Distance during reaction} = 20 , \mathrm{m/s} \times 0.4 , \mathrm{s} = 8 , \mathrm{m} $$
Thus, after his reaction, the distance between the car and the boy decreases to: $$ \text{Distance remaining} = 48 , \mathrm{m} - 8 , \mathrm{m} = 40 , \mathrm{m} $$
To stop the car within the remaining $40 , \mathrm{m}$, using the equation of motion: $$ v^2 = u^2 + 2as $$ where ( v = 0 , \mathrm{m/s} ) (final velocity), ( u = 20 , \mathrm{m/s} ) (initial velocity), and ( s = 40 , \mathrm{m} ) (distance to stop). Rewriting to find ( a ) (acceleration, which is actually the retardation here since we are stopping): $$ 0 = (20)^2 + 2 \times a \times 40 \ -400 = 80a \ a = \frac{-400}{80} = -5 , \mathrm{m/s}^2 $$
The negative sign indicates retardation (deceleration). Therefore, the minimum retardation required is $5 , \mathrm{m/s}^2$, to safely stop the car and save the boy.
A car is travelling at a speed of $20 \mathrm{~m}/{s}$. Evaluate the value of speed in $\mathrm{km}/{hr}$.
A) $48 \mathrm{~km}/{hr}$
B) $72 \mathrm{~km}/{hr}$
C) $36 \mathrm{~km}/{hr}$
D) $54 \mathrm{~km}/{hr}$
To convert the speed from meters per second ($\mathrm{m/s}$) to kilometers per hour ($\mathrm{km/hr}$), we use the conversion factors where 1 kilometer is 1000 meters, and 1 hour is 3600 seconds. The speed of the car is given as $20 \mathrm{~m/s}$.
The conversion process involves two steps:
Convert meters to kilometers.
Convert seconds to hours.
For the given speed: $$ 20 , \mathrm{m/s} = 20 \times \frac{1000 , \mathrm{m}}{1 , \mathrm{km}} \times \frac{1 , \mathrm{hr}}{3600 , \mathrm{s}} = 20 \times \frac{3600}{1000} , \mathrm{km/hr} $$
Simplifying this gives: $$ 20 \times 3.6 = 72 , \mathrm{km/hr} $$
Therefore, the car's speed is $72$ km/hr. The correct choice is Option B: $72 \mathrm{~km/hr}$.
A wheel of radius 30 cm has 8 spokes in it. It is mounted on a fixed axle and spins at 5 rev/s. If an arrow of length 24 cm is to be shot parallel to the axis of the wheel without hitting the rim or the spokes of the wheel, what minimum speed must the arrow have? (Ignore the effect of gravity).
A) 9.6 m/s
B) 4.8 m/s
C) 2.4 m/s
D) 19.2 m/s
The correct answer is A) 9.6 m/s.
Given Data:
Angular speed, $ \omega = 5 \text{ rev/s} $, which needs conversion to radians per second since one revolution is $2\pi$ radians: $$ \omega = 5 \times 2\pi = 10\pi \text{ rad/s}. $$
Radius of the wheel, $ r = 30 \text{ cm} = 0.30 \text{ m} $.
Objective: Calculate the minimum speed of the arrow that is required to pass through the wheel without hitting the rim or spokes.
Calculation:
The wheel has 8 spokes, hence it has 8 sectors with each subtending an angle of $\frac{2\pi}{8} = \frac{\pi}{4} \text{ rad}$.
Time taken for one sector to pass a fixed position: $$ t = \frac{\text{Angular width of one sector}}{\text{Angular speed}} = \frac{\pi/4}{10\pi} = \frac{1}{40} \text{ s}. $$
The arrow should pass through the width of the sector without hitting the spokes, so: $$ \text{Distance} = 2r = 2 \times 0.30 \text{ m} = 0.60 \text{ m} $$ (double the radius for the total diameter of the wheel).
Minimum speed of the arrow to pass it without interruption: $$ \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{0.60 \text{ m}}{\frac{1}{40} \text{ s}} = 24 \text{ m/s}. $$
However, the arrow's length is 24 cm, and not the full diameter needs to clear at once but only half its length plus the arrow (since it's shot parallel). Hence, the required distance to clear is: $$ \text{Required clearing distance} = \text{Radius} + \frac{\text{Length of the arrow}}{2} = 0.30 \text{ m} + 0.12 \text{ m} = 0.42 \text{ m}. $$ Thus, the corrected minimum speed calculation: $$ \text{Speed} = \frac{0.42 \text{ m}}{\frac{1}{40} \text{ s}} = 16.8 \text{ m/s}. $$
Thus, rounding this down based on given options or recalculating the precise sector passing time might be necessary to match options. Recalculating with sector distance leads us to reassess the angular calculations or review the provided options for potential error. As per the given solution, it presumes the distance aligned with calculation as $\frac{24 cm}{0.42 m} = 57.14$, showing part-miscalculation but it leads us back to: $$ \text{Speed} = \frac{0.24 \text{ m}}{\frac{1}{40} \text{ s}} = 9.6 \text{ m/s}. $$ Hence, the minimum speed of the arrow must be 9.6 m/s to avoid hitting the wheel's spokes.
A ring of radius $2$ m performs combined translational and rotational motion on a frictionless horizontal surface with an angular acceleration of $4$ rad/s$^{2}$ and the acceleration of its center, $a = 4$ m/s$^{2}$, as shown in the figure. Find the acceleration of point $D$.
(A) $8 \hat{\mathrm{i}}$ m/s$^{2}$
B $(4 \hat{i}+8 \hat{j})$ m/s$^{2}$
C $(4 \hat{i}-4 \hat{j})$ m/s$^{2}$
D $8 \hat{\mathrm{j}}$ m/s$^{2}$
The correct answer is $\mathbf{(4 \hat{i}+8 \hat{j}) \text{ m/s}^2}$.
Given that the radius of the ring, $r$, is $2$ m and the angular acceleration $\vec{\alpha} = -4 \hat{k}$ rad/s$^2$ (determined using the right-hand thumb rule), let's find the acceleration of point $D$ on the ring.
The acceleration of the center of the ring, denoted $\vec{a}_0$, is $4 \hat{i} \text{ m/s}^2$. To find the total acceleration at point $D$, $\vec{a}_D$, consider:
The translational acceleration which is $\vec{a}_0 = 4 \hat{i} \text{ m/s}^2$.
The rotational contribution to the acceleration at $D$, $\vec{a}{DO}$, which can be determined by: $$ \vec{a}{DO} = \vec{\alpha} \times \vec{r}{DO} $$ where $\vec{r}{DO} = -2 \hat{i} \text{ m}$ represents the position vector of point $D$ from the center.
Substituting the given values: $$ \vec{a}_{DO} = (-4 \hat{k}) \times (-2 \hat{i}) = 8 (\hat{k} \times \hat{i}) = 8 \hat{j} $$
Adding the translational and rotational parts: $$ \vec{a}_D = 4 \hat{i} + 8 \hat{j} $$
Thus, the acceleration of point $D$ is $\vec{a}_D = (4 \hat{i} + 8 \hat{j}) \text{ m/s}^2$.
Use the following information to answer the next question: Shivani throws a stone upward with a velocity of $19.6 \mathrm{~m/s}$. The acceleration of the stone in this case is $9.8 \mathrm{~m/s}^{2}$, acting toward Earth. What is the maximum height reached by the stone?
A) $9.8 \mathrm{~m}$
B) $14.2 \mathrm{~m}$
C) $19.6 \mathrm{~m}$
D) $28.4 \mathrm{~m}$
The correct answer is C) $19.6 \mathrm{~m}$.
To find the maximum height reached by the stone, let's use the kinematic equation since it's thrown vertically upward:
$$ v^2 = u^2 + 2as $$
Here:
$v$ is the final velocity at the highest point, which becomes $0 \mathrm{~m/s}$ (since the stone stops momentarily at the highest point before starting to descend),
$u$ is the initial velocity, $19.6 \mathrm{~m/s}$,
$a$ is the acceleration due to gravity, but since it acts downward while the stone moves upward, it becomes $-9.8 \mathrm{~m/s}^2$,
$s$ is the displacement or the maximum height, which we are required to determine.
Plugging in the values:
$$ 0 = (19.6)^2 + 2 \times (-9.8) \times s $$ $$ 0 = 384.16 - 19.6s $$ $$ 19.6s = 384.16 $$ $$ s = \frac{384.16}{19.6} = 19.6 \mathrm{~m} $$
Thus, the maximum height reached by the stone is $19.6$ meters.
A motorcycle's horn is blowing out a $3015 \mathrm{~Hz}$ sound as it moves eastward towards a stationary observer. A train sounds its horn at $855 \mathrm{~Hz}$ as it heads westward away from that same observer. If the observer hears the train as $850 \mathrm{~Hz}$ and the motorcycle's horn as $3060 \mathrm{~Hz}$, what is the speed of the motorcycle relative to the train (in $\mathrm{m/s}$)? Use $340 \mathrm{~m/s}$ as the speed of sound in air.
For the motorcycle, we start by applying the formula for the observed frequency when a source moves toward the observer: $$ f = f_0 \left(\frac{V}{V - V_S}\right) $$ Here, $f_0 = 3015 \text{ Hz}$ is the original frequency, $V = 340 \text{ m/s}$ is the speed of sound, and $f = 3060 \text{ Hz}$ is the observed frequency. We aim to find $V_S$, the speed of the motorcycle. Substituting the values, we get: $$ 3060 = 3015 \left(\frac{340}{340 - V_M}\right) $$ Rearranging to solve for $V_M$: $$ 340 - V_M = \frac{3015}{3060} \times 340 \Rightarrow V_M = 340 - \frac{3015}{3060} \times 340 \approx 5 \text{ m/s} $$
Next, for the train, the formula for the observed frequency when the source moves away from the observer: $$ f = f_0 \left(\frac{V}{V + V_S}\right) $$ Here, $f_0 = 855 \text{ Hz}$, and $f = 850 \text{ Hz}$. Substituting, $$ 850 = 855 \left(\frac{340}{340 + V_T}\right) $$ Solving for $V_T$: $$ 340 + V_T = \frac{340 \times 855}{850} \Rightarrow V_T = \frac{340 \times 855}{850} - 340 \approx 2 \text{ m/s} $$
Now, we find the relative speed between the motorcycle and the train: $$ V_{\text{relative}} = V_M - (-V_T) = 5 - (-2) = 7 \text{ m/s} $$
Therefore, the relative speed of the motorcycle with respect to the train is $7 \text{ m/s}$.
A block of mass '$m$' is placed on a triangular block of mass '$\mathrm{M}$', which in turn is placed on a horizontal smooth surface as shown in figure. Assuming frictionless surfaces, find the velocity of the triangular block when the smaller block reaches the bottom end.
(A) $\left[\frac{2 M^{2} g h \cos^{2} \alpha}{(m+M)(m+M \sin^{2} \alpha)}\right]^{\frac{1}{2}}$ (B) $\left[\frac{2 M^{2} g h \cos^{2} \alpha}{(M+m)(M+m \sin^{2} \alpha)}\right]^{\frac{1}{2}}$ (C) $\left[\frac{2 M^{2} g h \sin^{2} \alpha}{(M+m)(M+m \sin^{2} \alpha)}\right]^{\frac{1}{2}}$ (D) $\left[\frac{2 M^{2} g h \sin^{2} \alpha}{(m+M)(m+M \cos^{2} \alpha)}\right]^{\frac{1}{2}}$
Correct Option: $\mathbf{B}$
Velocity of the Triangular Block:$$ \left[\frac{2 M^2 g h \cos^2 \alpha}{(M+m)(M+m \sin^2 \alpha)}\right]^{\frac{1}{2}} $$
Explanation:
Relative Acceleration of Block $m$: The smaller block $m$ accelerates down the incline with acceleration $a_1 = g \sin \alpha$ due to gravity.
Horizontal Component of Acceleration ($a_x$): This acceleration can be resolved into a horizontal component given by: $$ a_x = g \sin \alpha \cos \alpha $$
Acceleration of Block $M$: The triangular block $M$ moves left owing to $a_x$. Let its acceleration be $a_2$. Using the conservation of momentum in the horizontal direction (since external horizontal force is zero): $$ m a_x = (M+m) a_2 \ \Rightarrow a_2 = \frac{m g \sin \alpha \cos \alpha}{M+m} $$
Net Acceleration of $m$ on $M$:$$ a = g \sin \alpha - a_2 \cos \alpha \ = g \sin \alpha - \frac{m g \sin \alpha \cos^2 \alpha}{M+m} \ = g \sin \alpha \left(1 - \frac{m \cos^2 \alpha}{M+m}\right) $$
Time to Reach the Bottom ($t$): The time it takes for block $m$ to reach the bottom can be found from: $$ \text{Distance along incline} = \frac{h}{\sin \alpha} = \frac{1}{2} a t^2 \ \Rightarrow t = \sqrt{\frac{2h}{a \sin \alpha}} $$
Final Velocity of Block $M$ ($V_M$):$$ V_M = a_2 t = \frac{m g \sin \alpha \cos \alpha}{M+m} \cdot \sqrt{\frac{2h}{a \sin \alpha}} \ = \left[\frac{2 M^2 g h \cos^2 \alpha}{(M+m)(M+m \sin^2 \alpha)}\right]^{\frac{1}{2}} $$
This derivation leads us to conclude that the velocity of the triangular block when the smaller block reaches the bottom is described by option B.
Which term refers to the movement of molecules from an area of higher concentration to an area of lower concentration?
A. active transport
B. diffusion
C. concentration
D. collision
The correct answer is B. diffusion.
Diffusion is the movement of molecules from an area of higher concentration to an area of lower concentration. This process occurs until there is an equilibrium in concentration levels across the space.
A car of mass $1500 \mathrm{~kg}$ travelling at $25 \mathrm{~m/s}$ collides with another car B of mass $1000 \mathrm{~kg}$ travelling at $15 \mathrm{~m/s}$ in the same direction. After the collision, the velocity of car B will be:
To find the velocity of car B after the collision, we apply the law of conservation of momentum. Given are:
Mass of car A, $ m_1 = 1500 , \text{kg} $
Mass of car B, $ m_2 = 1000 , \text{kg} $
Initial velocity of car A, $ u_1 = 25 , \text{m/s} $
Initial velocity of car B, $ u_2 = 15 , \text{m/s} $
Final velocity of car A after collision, $ v_1 = 20 , \text{m/s} $
Final velocity of car B after collision, $ v_2 = ? $
By the conservation of momentum: $$ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 $$ Substituting the values: $$ 1500 \times 25 + 1000 \times 15 = 1500 \times 20 + 1000 \times v_2 $$ This simplifies to: $$ 37500 + 15000 = 30000 + 1000 v_2 $$ $$ 52500 = 30000 + 1000 v_2 $$ Solving for $ v_2 $: $$ 22500 = 1000 v_2 $$ $$ v_2 = \frac{22500}{1000} = 22.5 , \text{m/s} $$
Thus, the velocity of car B after collision is $ 22.5 , \text{m/s} $.
$10 \mathrm{~m} \mathrm{~s}^{-1}$ is equal to $36 \mathrm{~km} \mathrm{~hr}^{-1}$.
A True
B False
Solution The correct answer is A True.
To convert $10 , \text{m/s}$ to $\text{km/hr}$, we use the conversion factors:
$1 , \text{km} = 1000 , \text{m}$ (thus, $1 , \text{m} = \frac{1}{1000} , \text{km}$)
$1 , \text{hr} = 3600 , \text{s}$ (thus, $1 , \text{s} = \frac{1}{3600} , \text{hr}$).
Then, convert $10 , \text{m/s}$ to $\text{km/hr}$ as follows: $$ 10 , \text{m/s} = \frac{10 , \text{m}}{1 , \text{s}} = \frac{10/1000 , \text{km}}{1/3600 , \text{hr}} = 36 , \text{km/hr} $$ Thus, $10 , \text{m/s}$ is indeed equal to $36 , \text{km/hr}$, confirming that the statement is true.
If the speed of a car is $5 \mathrm{~ms}^{-1}$, what is the speed in $\mathrm{km} \mathrm{h}^{-1}$?
A. 18
B. 17
C. 16
D. 15
The correct option is A. 18
Given: The speed of the car is $5 \mathrm{~ms}^{-1}$, which can be expressed as $\frac{5 \mathrm{~m}}{1 \mathrm{~s}}$.
We know that $1 \mathrm{~km} \mathrm{~h}^{-1}$ is equivalent to $\frac{1000 \mathrm{~m}}{3600 \mathrm{~s}}$, which simplifies to $\frac{5}{18} \mathrm{~ms}^{-1}$.
To convert from $\mathrm{ms}^{-1}$ to $\mathrm{km} \mathrm{~h}^{-1}$, we multiply by $\frac{18}{5}$. Thus, the calculation is as follows: $$ 5 \mathrm{~ms}^{-1} \times \frac{18}{5} = 18 \mathrm{~km} \mathrm{~h}^{-1} $$
Therefore, $5 \mathrm{~ms}^{-1}$ is equal to $18 \mathrm{~km} \mathrm{~h}^{-1}.
A football player kicked a ball with a velocity of $20 \mathrm{~ms}^{-1}$ in a direction making an angle of $60^{\circ}$ with the horizontal. Find the velocity with which he should run in the same direction so that the ball remains above his head.
A) $15 \mathrm{~m/s}$ B) $10 \mathrm{~m/s}$ C) $20 \mathrm{~m/s}$ D) $5 \mathrm{~m/s}$
The correct answer is Option B) $10 \mathrm{~m/s}$.
To solve this problem, we need to precisely determine the horizontal component of the ball's velocity post-kick. The ball is kicked at a speed of $20 \mathrm{~ms}^{-1}$ and forms a $60^\circ$ angle with the horizontal. The horizontal and vertical velocities can be resolved using trigonometric functions from this initial velocity.
Horizontal Component Calculation: By using the cosine trigonometric function for the horizontal component, we have: $$ v_{\text{ball horizontal}} = v_{\text{ball}} \cos \theta $$ where $\theta = 60^\circ$. Thus, substituting the values, $$ v_{\text{ball horizontal}} = 20 \cos 60^\circ = 20 \times \frac{1}{2} = 10 , \text{m/s} $$
Therefore, for the football player to keep the ball vertically aligned above him as he moves, he needs to run at the horizontal velocity of the ball, which is $10 , \text{m/s}$.
A cord is wrapped around a pulley that is shaped like a disk of mass $m$ and radius $r$. The cord's free end is connected to a block of mass $M$. The block starts from rest and then slides down an incline that makes an angle $\theta$ with the horizontal as shown in the figure. The coefficient of kinetic friction between the block and the incline is $\mu$. Then the magnitude of the acceleration of the block is:
A) $\frac{2Mg(\sin\theta-\mu\cos\theta)}{2M+m}$
B) $\frac{2Mg(\sin\theta+\mu\cos\theta)}{2M+m}$
C) $\frac{2Mg(\sin\theta-\mu\cos\theta)}{2M-m}$
D) $\frac{Mg(\sin\theta-\mu\cos\theta)}{2M+m}$
Solution: The correct option is A: $$ \frac{2Mg(\sin\theta - \mu\cos\theta)}{2M+m} $$
Analyzing the forces and torques:
For the block of mass $M$, considering the forces along the direction of the incline: $$ Mg\sin\theta - T - \mu Mg\cos\theta = Ma $$ where $T$ is the tension in the cord.
For the pulley, the net torque $\tau$ generated by the tension and the moment of inertia $I$ of the pulley (a disk) gives: $$ Tr = I\alpha $$ For a disk, $I = \frac{mr^2}{2}$ and $\alpha = \frac{a}{r}$ (since $\alpha r = a$ for rolling without slipping), thus: $$ Tr = \frac{mr^2}{2} \cdot \frac{a}{r} $$ $$ T = \frac{ma}{2} $$
Combining equations:Substituting $T$ into the block's equation: $$ Mg\sin\theta - \frac{ma}{2} - \mu Mg\cos\theta = Ma $$ Rearranging for $a$: $$ a = \frac{2Mg(\sin\theta - \mu\cos\theta)}{2M+m} $$
This leads us to the correct answer, which incorporates the block’s sliding down an incline with friction and the influence of both the mass and the rotational inertia of the pulley.
A train covers $54 \mathrm{~km}$ in 10 minutes. Find the speed of the train.
(A) $60 \mathrm{~m} \mathrm{~s}^{-1}$
(B) $70 \mathrm{~m} \mathrm{~s}^{-1}$
(C) $80 \mathrm{~m} \mathrm{~s}^{-1}$
(D) $90 \mathrm{~m} \mathrm{~s}^{-1}$
To find the speed of the train, we start by converting the given parameters into compatible units.
Given:
Distance covered by the train: $d = 54 \text{ km}$
Convert kilometers to meters: $$ d = 54 \times 1000 \text{ m} = 54000 \text{ m} $$
Time taken: $t = 10 \text{ min}$
Convert minutes to seconds: $$ t = 10 \times 60 \text{ s} = 600 \text{ s} $$
Calculation:
The average speed ($v$) is defined as the total distance covered divided by the total time taken.
Calculating speed: $$ v = \frac{d}{t} = \frac{54000 \text{ m}}{600 \text{ s}} = 90 \text{ m/s} $$
Conclusion:
The speed of the train is 90 m/s, which corresponds to option (D) $90 \text{ m/s}$.
A marble with speed $20 \mathrm{~cm/s}$ rolls off the edge of a table $80 \mathrm{~cm}$ high. How long does it take to drop to the floor? How far horizontally from the table edge does the marble strike the floor?
A) $8.4 \mathrm{~m}$
B) $8.0 \mathrm{~cm}$
C) $8.0 \mathrm{~m}$
D) $8.9 \mathrm{~cm}$
Here's the complete solution to find how long it takes for a marble to drop to the floor and how far it lands from the edge:
Initial horizontal and vertical velocities:
The horizontal initial velocity of the marble is given as $$ v_{ox} = 20 , \text{cm/s} $$ and the initial vertical velocity is zero, since the marble rolls off horizontally: $$ v_{oy} = 0 $$Acceleration due to gravity: Since the vertical motion is affected by gravity, the acceleration in the vertical direction ($a_y$) is the acceleration due to gravity: $$ a_y = +g = +980 , \text{cm/s}^2 $$ Note that the horizontal acceleration ($a_x$) is zero as there is no horizontal force acting on the marble: $$ a_x = 0 $$
Time to hit the floor: Using the equation for vertical motion, where displacement ($y$) is given as the negative height of the table (because it is falling down), $$ y = v_{oy} t + \frac{1}{2} a_y t^2 $$ Plugging in the values, $$ -80 , \text{cm} = 0 + \frac{1}{2} \times 980 , \text{cm/s}^2 \times t^2 $$ Solving for $t$, we get: $$ t^2 = \frac{160}{980} \approx 0.16326 , \text{s}^2\ t \approx 0.40 , \text{s} $$
Horizontal distance: The horizontal distance $x$ it travels is computed by: $$ x = v_{ox} t = 20 , \text{cm/s} \cdot 0.40 , \text{s} = 8.0 , \text{cm} $$
Therefore, the marble takes approximately 0.40 seconds to hit the floor and lands 8.0 cm away from the base of the table. The correct choice is B) $8.0 , \text{cm}$.
A truck has a mass of 8000 kgs, a car has a mass of 1000 kgs, and a bike has a mass of 100 kgs. Both the truck and car are moving on a level road at 45 ms^-1. The bike, however, is going on the same level road at 55 ms^-1. The kinetic energy will be more for the:
A truck
B car
C bike
D truck and bike only
The correct option is A) truck.
Kinetic Energy (KE) is fundamentally influenced by both mass and velocity, following the formula: $$ KE = \frac{1}{2} m v^2 $$ where:
$m$ is the mass of the object,
$v$ is the velocity of the object.
Here, the truck and the car both travel at the same speed of $45 , \text{ms}^{-1}$, and the bike is faster at $55 , \text{ms}^{-1}$. Despite the bike’s faster speed, its significantly lower mass compared to the truck (100 kgs versus 8000 kgs) leads to a lower kinetic energy. Therefore, among the options, the truck, having the highest mass and a decent speed, possesses the highest kinetic energy.
9 What is the momentum of an object of mass $m$, moving with a velocity $v$? (a) $(mv)^{2}$ (b) $mv^{2}$ (c) $\frac{1}{2} mv^{2}$ (d) $mv$
Solution
The correct answer is (d) $ mv $.
Momentum is defined as the product of an object's mass ($ m $) and its velocity ($ v $). Therefore, the momentum $ p $ of an object can be calculated using the formula: $$ p = m \times v $$ where
$ m $ represents the mass of the object,
$ v $ represents the velocity of the object.
Thus, the momentum $ p $ is expressed as: $$ p = mv $$
So, the correct choice is (d) $ mv $.
A ball is thrown upwards with a velocity of $10 \mathrm{~ms}^{-1}$. Find the maximum height reached by the ball.
A) $13 \mathrm{~m}$
B) $5 \mathrm{~m}$
C) $15 \mathrm{~m}$
D) $16 \mathrm{~m}$
Solution
The correct option is B) $5 \mathrm{~m}$.
Given:
Initial velocity, $u = 10 \mathrm{~ms}^{-1}$
Acceleration due to gravity is, $g \approx -10 \mathrm{~m/s}^2$. *Note: The upward direction is considered positive.*
At the highest point, the velocity of the ball becomes zero since it changes its direction of motion: $$ v = 0 \mathrm{~ms}^{-1}. $$
Let the maximum height reached be denoted as $H$.
Applying the third equation of motion: $$ v^2 = u^2 + 2as $$ Substitute $v = 0$, $u = 10 \mathrm{~ms}^{-1}$, $a = -10 \mathrm{~m/s}^2$, and solve for $s$ (which is $H$ here): $$ 0 = 10^2 + 2(-10)H \ 100 = -20H \ H = \frac{100}{-20} \ H = 5 \mathrm{~m}. $$
Thus, the maximum height reached by the ball is $5 \mathrm{~m}$.
A body 'A' starts from rest with an acceleration $a_{1}$. After 2 seconds, another body 'B' starts from rest with an acceleration $a_{2}$. If they travel equal distances in the $5^{\text{th}}$ second after the start of $A$, then the ratio $a_{1}: a_{2}$ is equal to
A) $5: 9$
B) $5: 7$
C) $9: 5$
D) $9: 7$
The correct answer is A) $5:9$.
To find the distance traveled by each body during a specific second, we use the formula for the distance covered during the $n$-th second: $$ S_{n\text{th}} = u + \frac{a}{2}(2n - 1) $$ where $u$ is the initial velocity and $a$ is the acceleration.
Since both bodies start from rest, their initial velocities are zero: $u_A = u_B = 0$.
For body $A$, which starts moving $2$ seconds before $B$, during its $5$-th second: $$ n_A = 5 $$
For body $B$, which starts moving from rest after $A$ has already been moving for $2$ seconds, during its $3$-rd second: $$ n_B = 3 $$
Given that the distances traveled by both bodies during these respective seconds are equal, we set these distances to each other: $$ \frac{a_1}{2}(2 \times 5 - 1) = \frac{a_2}{2}(2 \times 3 - 1) $$
Simplifying both sides of the equation, we have: $$ \frac{a_1}{2}(9) = \frac{a_2}{2}(5) $$
Dividing through by $1/2$ and solving for the ratio $\frac{a_1}{a_2}$: $$ 9a_1 = 5a_2 \ \frac{a_1}{a_2} = \frac{5}{9} $$
Thus, the ratio $a_1:a_2$ is $5:9$, as needed.
A water tanker filled up to two-thirds of its height is moving with a uniform speed. On sudden application of brake, the water in the tank will:
move backward
move forward
rise up
remain unaffected
Correct option: B: move forward
When a tanker filled up to two-thirds of its height is moving with a uniform speed and then suddenly comes to a stop due to the application of brakes, the water inside the tank exhibits inertia of motion. This means that while the tanker comes to a halt, the water inside it continues to move forward owing to its inertia. Consequently, the water in the tank will move forward before eventually settling down.
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Ask Chatterbot AINCERT Solutions - Motion | NCERT | Science | Class 9
An athlete completes one round of a circular track of diameter $200 \mathrm{~m}$ in $40 \mathrm{~s}$. What will be the distance covered and the displacement at the end of 2 minutes $20 \mathrm{~s}$ ?
To find the distance covered and the displacement at the end of 2 minutes and 20 seconds by the athlete on a circular track with a diameter of $200 , \text{m}$, we need to calculate the following:
Circumference of the track: The circumference $ C $ of a circle with diameter $ d $ is given by the formula: $ C = \pi d $
Number of rounds in 2 minutes and 20 seconds: First, convert the total time from minutes and seconds to seconds:
$$ 2 \text{ min} \times 60 \text{ sec/min} + 20 \text{ sec} = 120 \text{ sec} + 20 \text{ sec} = 140 \text{ sec} $$
Next, calculate the number of rounds $ N $ using the time taken for one round $ T $ (which is 40 seconds):
$$ N = \frac{\text{Total Time}}{T} $$
Total distance covered: The total distance $ D $ is the number of rounds multiplied by the circumference: $ D = N \times C $
Displacement at the end of 2 minutes and 20 seconds: The displacement after completing full rounds is zero, as the athlete will be back at the starting point. If there is a fraction of the last round remaining, the displacement is the straight-line distance from the starting point to the point where the athlete stops. If the remaining distance corresponds to less than half a round, the displacement is the chord of the circle, which can be calculated using the angle traversed on the last incomplete round.
Let’s proceed with these calculations.
The circumference ( C ) of the circle is:
$$ C = \pi d = \pi \times 200 , \text{m} $$
The number of full rounds $ N $ completed in $ 140 , \text{sec} $ is:
$$ N = \frac{140}{40} $$
The number of full rounds is the integer part of $ N $, whereas any fractional part would represent an incomplete round.
Let's start by calculating the circumference $ C $ and the number of rounds $ N $.
The circumference ( C ) of the track is:
$$ C = \pi \cdot d = \pi \cdot 200 \text{ m} $$
Since $ \pi $ is approximately $ 3.14159 $, we have:
$$ C \approx 3.14159 \cdot 200 \text{ m} = 628.318 \text{ m} $$
Rounded to three significant figures, the circumference is about $ 628 \text{ m} $.
The number of rounds ( N ) is given by:
$$ N = \frac{140 \text{ s}}{40 \text{ s/round}} = 3.5 \text{ rounds} $$
The athlete completes $ 3 $ full rounds and half of another round in $ 140 \text{ s} $.
The distance covered, which is the total path length traveled, is: $ D = N \cdot C = 3.5 \cdot 628 \text{ m} \approx 2198 \text{ m} $
The displacement is the straight-line distance from the start to the end point. Since the athlete covers an exact number of full rounds plus a half round, the displacement is equal to the diameter of the circular track, which is $ 200 \text{ m} $.
Therefore, the distance covered is approximately $ 2198 \text{ m} $, and the displacement is $ 200 \text{ m} $.
Joseph jogs from one end A to the other end B of a straight $300 \mathrm{~m}$ road in 2 minutes 30 seconds and then turns around and jogs $100 \mathrm{~m}$ back to point $\mathrm{C}$ in another 1 minute. What are Joseph's average speeds and velocities in jogging (a) from $\mathrm{A}$ to $\mathrm{B}$ and (b) from $\mathrm{A}$ to $\mathrm{C}$ ?
To calculate Joseph's average speed and velocity for both parts (a) from A to B and (b) from A to C, we need to use the following definitions:
Average Speed is the total distance traveled divided by the total time taken.
Average Velocity is the displacement (change in position) divided by the total time taken.
Let's start with part (a): from A to B.
(a) From A to B:
Distance traveled from A to B: $300 \, \mathrm{m}$
Time taken to jog from A to B: $2 \, \mathrm{minutes} , 30 \, \mathrm{seconds} = 2 \times 60 \, \mathrm{seconds} + 30 \, \mathrm{seconds} = 150 \, \mathrm{seconds}$
Average Speed from A to B:
$$ \text{Average Speed}_{\text{A to B}} = \frac{\text{Distance}_{\text{A to B}}}{\text{Time taken}_{\text{A to B}}} = \frac{300 \, \mathrm{m}}{150 \, \mathrm{s}} $$
The average velocity from A to B will be the same as the average speed because the displacement in this straight path is equal to the distance covered.
Average Velocity from A to B:
$$ \text{Average Velocity}_{\text{A to B}} = \frac{\text{Displacement}_{\text{A to B}}}{\text{Time taken}_{\text{A to B}}} = \frac{300 \, \mathrm{m}}{150 \, \mathrm{s}} $$
Now, let's calculate both the average speed and the average velocity for part (a) from A to B:
Average Speed from A to B:
$$ \text{Average Speed}_{\text{A to B}} = \frac{300 \, \mathrm{m}}{150 \, \mathrm{s}} = 2 \, \mathrm{m/s} $$
Average Velocity from A to B:
$$ \text{Average Velocity}_{\text{A to B}} = \frac{300 \, \mathrm{m}}{150 \, \mathrm{s}} = 2 \, \mathrm{m/s} $$
Now we move on to part (b).
(b) From A to C:
Distance traveled from A to C is the sum of the distance from A to B and back to C: $300 \, \mathrm{m} + 100 \, \mathrm{m} = 400 \, \mathrm{m}$
Displacement from A to C is the distance from A to B minus the distance from B to C: $300 \, \mathrm{m} - 100 \, \mathrm{m} = 200 \, \mathrm{m}$
Total time taken to jog from A to C: $2 \, \mathrm{minutes} , 30 \, \mathrm{seconds} + 1 \, \mathrm{minute} = 150 \, \mathrm{seconds} + 60 \, \mathrm{seconds} = 210 \, \mathrm{seconds}$
Average Speed from A to C:
$$ \text{Average Speed}_{\text{A to C}} = \frac{\text{Distance}_{\text{A to C}}}{\text{Time taken}_{\text{A to C}}} = \frac{400 \, \mathrm{m}}{210 \, \mathrm{s}} $$
Average Velocity from A to C:
$$ \text{Average Velocity}_{\text{A to C}} = \frac{\text{Displacement}_{\text{A to C}}}{\text{Time taken}_{\text{A to C}}} = \frac{200 \, \mathrm{m}}{210 \, \mathrm{s}} $$
Now, let's calculate both the average speed and the average velocity for part (b) from A to C:
Average Speed from A to C:
$$ \text{Average Speed}_{\text{A to C}} \approx \frac{400 \, \mathrm{m}}{210 \, \mathrm{s}} \approx 1.905 \, \mathrm{m/s} $$
Average Velocity from A to C:
$$ \text{Average Velocity}_{\text{A to C}} \approx \frac{200 \, \mathrm{m}}{210 \, \mathrm{s}} \approx 0.952 \, \mathrm{m/s} $$
In conclusion, Joseph's average speed and velocity from A to B is $2$ m/s, and his average speed from A to C is approximately $1.905$ m/s with an average velocity of approximately $0.952$ m/s.
Abdul, while driving to school, computes the average speed for his trip to be $20 \mathrm{~km} \mathrm{~h}^{-1}$. On his return trip along the same route, there is less traffic and the average speed is $30 \mathrm{~km} \mathrm{~h}^{-1}$. What is the average speed for Abdul's trip?
Abdul's average speed for the entire trip cannot be simply calculated as the average of the two speeds (20 km/h and 30 km/h) because the time spent traveling at each speed is not the same — unless the two trips took exactly the same amount of time, which is not the case here due to different speeds. Instead, we need to use the definition of average speed:
$$ \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} $$
Let's assume the distance to the school (one way) is $ D $ kilometers. Abdul makes two trips, so the total distance $ d_{\text{total}} $ is:
$$ d_{\text{total}} = 2D $$
Now we need to calculate the time taken for each part of the trip. The time taken to go to school, $ t_{1} $, at 20 km/h for a distance $ D $ is:
$$ t_{1} = \frac{D}{20 \text{ km/h}} $$
And the time taken to return, $ t_{2} $, at 30 km/h for the same distance $ D $ is:
$$ t_{2} = \frac{D}{30 \text{ km/h}} $$
The total time $ t_{\text{total}} $ is the sum of $ t_{1} $ and $ t_{2} $:
$$ t_{\text{total}} = t_{1} + t_{2} = \frac{D}{20 \text{ km/h}} + \frac{D}{30 \text{ km/h}} $$
Now to find the average speed, $ v_{\text{avg}} $, for the entire trip, we use $ d_{\text{total}} $ and $ t_{\text{total}} $:
$$ v_{\text{avg}} = \frac{d_{\text{total}}}{t_{\text{total}}} = \frac{2D}{\frac{D}{20 \text{ km/h}} + \frac{D}{30 \text{ km/h}}} $$
Let's calculate this. The distance $ D $ cancels out, and we don't need to know its exact value:
$$ v_{\text{avg}} = \frac{2}{\frac{1}{20} + \frac{1}{30}} = \frac{2}{\frac{3}{60} + \frac{2}{60}} = \frac{2}{\frac{5}{60}} $$
$$ v_{\text{avg}} = 2 \times \frac{60}{5} = 2 \times 12 = 24 \text{ km/h} $$
So the average speed for Abdul's entire trip is 24 km/h.
A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of $3.0 \mathrm{~m} \mathrm{~s}^{-2}$ for $8.0 \mathrm{~s}$. How far does the boat travel during this time?
To determine the distance traveled by the motorboat during its acceleration, we can use the kinematic equation for uniformly accelerated motion. This equation relates the distance $s$, initial velocity $u$, acceleration $a$, and time $t$:
$$ s = ut + \frac{1}{2}at^2 $$
In this case, the boat is starting from rest, so the initial velocity $u$ is $0$. Thus, the equation simplifies to:
$$ s = \frac{1}{2}at^2 $$
Given the acceleration $a = 3.0 \mathrm{~m} \mathrm{~s}^{-2}$ and time $t = 8.0 \mathrm{~s}$, we can calculate the distance $s$:
$$ s = \frac{1}{2}(3.0 \mathrm{~m} \mathrm{~s}^{-2})(8.0 \mathrm{~s})^2 $$
Substituting the values and calculating:
$$ s = \frac{1}{2}(3.0)(64.0) = \frac{1}{2}(192.0) = 96.0 \mathrm{~m} $$
Therefore, the boat travels $ 96.0 \mathrm{~m} $ during the 8 seconds of acceleration.
Fig 7.10 shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions:
(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?
(c) How far has $\mathrm{C}$ travelled when $\mathrm{B}$ passes $A$?
(d) How far has B travelled by the time it passes C?
(a) The slope of the line on a distance-time graph represents the speed of the object. The steeper the slope, the faster the object is moving. Object B has the steepest slope among the three, so B is traveling the fastest.
(b) On a distance-time graph, the point where the lines intersect would indicate that the objects are at the same point on the road at the same time. The lines for objects A, B, and C do not all intersect at a single point, so they are not at the same point on the road at any time.
(c) B passes A at the point where their lines intersect, at approximately 1 hour. By then $C$ has travelled $1 \mathrm{~km}$
(d) By the time $B$ passes $C$ it has travelled approximately 9 km.
A ball is gently dropped from a height of $20 \mathrm{~m}$. If its velocity increases uniformly at the rate of $10 \mathrm{~m} \mathrm{~s}^{-2}$, with what velocity will it strike the ground? After what time will it strike the ground?
To solve this problem, we can apply the equations of motion for constant acceleration. Since the ball is dropped from rest, its initial velocity $ u $ is 0 m/s. Let's denote the final velocity with which the ball strikes the ground as $ v $, the acceleration due to gravity as $ a = 10 \text{ m/s}^2 $ (which is equal to the given uniform rate of increase in velocity), and the height from which the ball is dropped as $ h = 20 \text{ m} $.
Using the second equation of motion which relates final velocity, initial velocity, acceleration, and distance traveled: $ v^2 = u^2 + 2ah $
Substitute the given values:
$$ v^2 = 0^2 + 2 \cdot 10 \cdot 20 $$
$$ v^2 = 400 $$
Taking the square root (since velocity is positive for downward motion):
$$ v = \sqrt{400} = 20 \text{ m/s} $$
So, the ball will strike the ground with a velocity of 20 m/s.
Next, using the first equation of motion to find the time $ t $ it takes for the ball to strike the ground:
$$ v = u + at $$
Again substituting the known values and the value of $ v $ we just calculated:
$$ 20 = 0 + 10 \cdot t $$
Solving for $ t $:
$$ t = \frac{20}{10} = 2 \text{ s} $$
The ball will strike the ground after 2 seconds.
The speed-time graph for a car is shown is Fig. 7.11.
(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?
(a) To find the distance travelled by the car over a period of time on a speed-time graph, you need to calculate the area under the graph between the start and end times.
Looking at the graph, you can see that the car's speed is not constant, so the area under the graph from 0 to 4 seconds will be a circular arc.
Let us assume that it is $\frac{1}{4}^\mathrm{th}$ of a circle with radius as $4 \mathrm{units}$
Then, the distance covered by the car will be:
$$ A = \frac{1}{4} \times \pi \times (4)^2$$
$$ A = 4\pi $$
which is approximately $12.56 \mathrm{~m}$
(b) Since the speed-time graph becomes a straight line after almost 6 seconds, the part of the graph after 6 seconds represent uniform motion of the car.
State which of the following situations are possible and give an example for each of these:
(a) an object with a constant acceleration but with zero velocity
(b) an object moving with an acceleration but with uniform speed.
(c) an object moving in a certain direction with an acceleration in the perpendicular direction.
(a) An object with a constant acceleration but with zero velocity is possible. This situation occurs at the very instant when an object that has been moving in one direction starts to move in the opposite direction, just as its velocity reaches zero before accelerating again. A simple example is a ball thrown straight up into the air: at the peak of its path, the ball's velocity is zero, but it is still experiencing a constant acceleration downwards due to gravity, $ g \approx 9.8 , \text{m/s}^2 $.
(b) An object moving with acceleration but with uniform speed is also possible. This scenario can happen when an object is moving in a circular path at a constant speed; although its speed (the magnitude of its velocity) does not change, its direction does. Therefore, it has a centripetal acceleration directed towards the center of the circle. A good example is a car moving at a constant speed around a circular track; the direction of the car's velocity is constantly changing, so it accelerates towards the center of the circle, but its speed remains constant.
(c) An object moving in a certain direction with an acceleration in the perpendicular direction can happen. This situation is typical for objects in uniform circular motion. For example, the Moon orbiting the Earth has a tangential velocity around the Earth, and it experiences an acceleration towards Earth (gravity) that is perpendicular to its velocity. Despite its constant speed along the circular path, it is always accelerating toward the center of the Earth, perpendicular to its direction of motion.
An artificial satellite is moving in a circular orbit of radius $42250 \mathrm{~km}$. Calculate its speed if it takes 24 hours to revolve around the earth.
To calculate the speed of an artificial satellite moving in a circular orbit, we can use the formula for the circumference of a circle, which represents the path the satellite travels during one full orbit.
The formula for the circumference ($ C $) of a circle is: $ C = 2\pi r $ where ($ r $) is the radius of the circle.
Given that the radius of the satellite's orbit is ($ 42250 \ \text{km} $), we can calculate the circumference as follows: $ C = 2\pi(42250 \ \text{km}) $
The time ($ T $) it takes for the satellite to make one full revolution is 24 hours. To find the speed ($ v $), we use the formula: $ v = \frac{\text{Distance}}{\text{Time}} $
For the satellite, the Distance is the circumference of its orbit, and the Time is its orbital period. Thus: $ v = \frac{C}{T} $
Converting 24 hours to seconds (since speed is often expressed in kilometers per second) gives us: $ T = 24 \ \text{hours} \times 3600 \ \frac{\text{seconds}}{\text{hour}} = 86400 \ \text{seconds} $
Now we can calculate the speed: $ v = \frac{2\pi \times 42250 \ \text{km}}{86400 \ \text{seconds}} $
Let's calculate the speed ($ v $). The circumference of the satellite's orbit is approximately ($ 265464.579 \ \text{km} $), and the time for one revolution, converted to seconds, is ($ 86400 \ \text{seconds} $).
Now we can calculate the speed ($ v $): $ v = \frac{265464.579 \ \text{km}}{86400 \ \text{seconds}} \approx 3.073 \ \text{km/s} $
Therefore, the speed of the artificial satellite in its circular orbit is approximately ($ 3.073 \ \text{km/s} $).
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Ask Chatterbot AINotes - Motion | Class 9 NCERT | Science
Everything You Need to Know
Understanding Motion in Physics (Class 9)
Motion is a fundamental concept in physics that describes the change in position of an object over time. From the flight of birds to the movement of planets in their orbits, motion is all around us and understanding it is crucial for exploring the physical world.
Different types of motion: A bird in flight, a car moving, and a planet's orbit.
1. What is Motion?
Defining Motion
Motion refers to the change in the position of an object with respect to time. For example, a car driving down the road, the blood flowing through veins, or planets orbiting a star are all instances of motion.
Types of Motion
Motion can be classified into various types:
Linear Motion: Movement in a straight line.
Circular Motion: Motion along a circular path.
Rotational Motion: When an object rotates around an axis.
Vibrational Motion: Rapid back and forth movement.
2. Describing Motion
Describing Motion
To describe motion effectively, it is essential to choose a reference point. For instance, if a school is 2 km north of a railway station, the railway station acts as the reference point.
3. Uniform and Non-Uniform Motion
Uniform Motion
Uniform motion occurs when an object covers equal distances in equal intervals of time. For example, riding a bike at a constant speed along a straight path.
Non-Uniform Motion
Non-uniform motion happens when an object covers unequal distances in equal intervals of time, such as a car navigating through traffic.
4. Measuring the Rate of Motion
Measuring Speed and Velocity
Speed is the distance covered per unit time and can be measured in meters per second (m/s). Velocity includes both speed and direction, making it a vector quantity.
Calculating Average Speed
Average speed can be calculated using the formula: [ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} ]
Speed vs. Velocity
While speed is scalar (only magnitude), velocity is a vector, meaning it has both magnitude and direction.
5. Understanding Acceleration
Acceleration in Motion
Acceleration is the rate of change of velocity. It can be calculated using: [ \text{Acceleration} (a) = \frac{\text{Change in Velocity}}{\text{Time Taken}} ]
Uniform vs. Non-Uniform Acceleration
Uniform Acceleration: Constant change in velocity over time.
Non-Uniform Acceleration: Variable change in velocity over time.
6. Graphical Representation of Motion
Distance-Time Graphs
Distance-time graphs can illustrate both uniform and non-uniform motion. In a uniform motion graph, the distance increases linearly with time, forming a straight line.
A distance-time graph illustrating uniform and non-uniform motion.
Velocity-Time Graphs
Velocity-time graphs represent the change in velocity over time. A straight horizontal line indicates uniform velocity, while a sloping line indicates acceleration.
7. Equations of Motion
Understanding Equations of Motion
For objects moving with uniform acceleration, the three primary equations of motion are used:
( v = u + at )
( s = ut + \frac{1}{2}at^2 )
( v^2 = u^2 + 2as )
Where ( u ) is the initial velocity, ( v ) is the final velocity, ( a ) is the acceleration, ( t ) is the time, and ( s ) is the distance.
8. Uniform Circular Motion
Exploring Uniform Circular Motion
Uniform circular motion occurs when an object moves in a circular path at a constant speed. Though the speed is constant, the direction changes continuously, implying acceleration.
Practical Examples and Activities
Examples include the motion of a satellite around Earth and a car making a turn on a circular track.
Conclusion
Key Takeaways from Motion Class 9 Notes
Understanding motion is critical for comprehending various physical phenomena. By mastering concepts such as speed, velocity, and acceleration, as well as utilizing graphical tools and equations of motion, students can enhance their grasp of physics.
FAQs
Frequently Asked Questions
What is the difference between speed and velocity?
Speed is a scalar quantity that only measures magnitude, while velocity is a vector quantity that includes both magnitude and direction.
What does the odometer of an automobile measure?
An odometer measures the distance traveled by the vehicle.
Remember, a thorough understanding of motion is foundational for physics. By studying the various types of motion and their representations, you can deepen your knowledge and prepare effectively for exams.
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