Multiple Choice Questions Solutions
2.1 The numerical ratio of displacement to distance is:
(a) Always less than one
(b) Always equal to one
(c) Always greater than one
(d) Equal to or less than one
Correct Option: (d) Equal to or less than one
Reason
Displacement is the shortest distance between two points, so it can be equal to or less than the actual distance traveled, but never greater.
2.2 If a body does not change its position with respect to some fixed point, then it will be in a state of:
(a) Rest
(b) Motion
(c) Uniform motion
(d) Variable motion
Correct Option: (a) Rest
Reason
If a body does not change its position relative to a fixed point, it is not moving, hence it is in a state of rest.
2.3 A ball is dropped from the top of a tower, the distance covered by it in the first second is:
(a) \( 5 \, \text{m} \)
(b) \( 10 \, \text{m} \)
(c) \( 50 \, \text{m} \)
(d) \( 100 \, \text{m} \)
Correct Option: (a) \( 5 \, \text{m} \)
Reason
Using the equation \( s = \frac{1}{2} g t^2 \), where \( g = 10 \, \text{m/s}^2 \) and \( t = 1 \, \text{s} \):
\[s = \frac{1}{2} (10) (1^2) = 5 \, \text{m}\]
2.4 A body accelerates from rest to a velocity of \( 144 \, \text{km/h} \) in \( 20 \, \text{s} \). Then the distance covered by it is:
(a) \( 100 \, \text{m} \)
(b) \( 400 \, \text{m} \)
(c) \( 1400 \, \text{m} \)
(d) \( 1440 \, \text{m} \)
Correct Option: (b) \( 400 \, \text{m} \)
Reason
First, convert \( 144 \, \text{km/h} \) to \( \text{m/s} \):
\[144 \, \text{km/h} = \frac{144 \times 1000}{3600} = 40 \, \text{m/s}\]
Using the formula \( s = \frac{1}{2} v t \):
\[s = \frac{1}{2} (40) (20) = 400 \, \text{m}\]
2.5 A body is moving with constant acceleration starting from rest. It covers a distance \( S \) in \( 4 \, \text{s} \). How much time does it take to cover one-fourth of this distance?
(a) \( 1 \, \text{s} \)
(b) \( 2 \, \text{s} \)
(c) \( 4 \, \text{s} \)
(d) \( 16 \, \text{s} \)
Correct Option: (b) \( 2 \, \text{s} \)
Reason
Using the equation \( s \propto t^2 \):
For one-fourth the distance, the time will be \( \sqrt{\frac{1}{4}} = \frac{1}{2} \) of the original time.
\[t = \frac{1}{2} \times 4 = 2 \, \text{s}\]
2.6 The displacement-time graphs of two objects A and B are shown in the figure. Point out the true statement from the following:
(a) The velocity of A is greater than B.
(b) The velocity of A is less than B.
(c) The velocity of A is equal to that of B.
(d) The graph gives no information in this regard.
Correct Option (a) The velocity of A is greater than B.
Reason
The slope of the displacement-time graph represents velocity. A has a steeper slope than B, so its velocity is greater.
2.7 The area under the speed-time graph is numerically equal to:
(a) Velocity
(b) Uniform velocity
(c) Acceleration
(d) Distance covered
Correct Option: (d) Distance covered
Reason
The area under the speed-time graph represents the total distance covered, as it is the integral of speed over time.
2.8 Gradient of the speed-time graph is equal to:
(a) Speed
(b) Velocity
(c) Acceleration
(d) Distance covered
Correct Option: (c) Acceleration
Reason
The gradient of a speed-time graph represents the rate of change of speed with respect to time, which is defined as acceleration.
2.9 Gradient of the distance-time graph is equal to the:
(a) Speed
(b) Velocity
(c) Distance covered
(d) Acceleration
Correct Option : (b) Velocity
Reason
The gradient of a distance-time graph shows the rate of change of distance with respect to time, which is velocity.
2.10 A car accelerates uniformly from \( 80.5 \, \text{km/h} \) at \( t = 0 \) to \( 113 \, \text{km/h} \) at \( t = 9 \, \text{s} \). Which graph best describes the motion of the car?
Which graph best describes the motion of the car?
Options are provided as graphs:
(a) Gradual increase and then sharp rise.
(b) Uniformly increasing straight-line graph.
(c) Non-linear increase.
(d) Linearly decreasing graph.
Correct Option:(b) Uniformly increasing straight-line graph
Reason
Since the car is accelerating uniformly, the velocity increases linearly with time, which corresponds to a straight-line graph with a positive slope.
Short Questions
2.1 Define scalar and vector quantities.
Scalar Quantities:
These are quantities that have only magnitude (size) and no direction. Examples include mass, time, and speed.
Vector Quantities:
These are quantities that have both magnitude and direction. Examples include force, velocity, and displacement.
2.2 Give 5 examples each for scalar and vector quantities.
Scalar Quantities
Mass
Time
Speed
Temperature
Distance
Vector Quantities
Velocity
Acceleration
Force
Displacement
Momentum
2.3 State head-to-tail rule for addition of vectors.
The head-to-tail rule states that to add two vectors, place the tail of the second vector at the head of the first vector. The resultant vector is drawn from the tail of the first vector to the head of the second vector.
2.4 What are distance-time graph and speed-time graph?
Distance-Time Graph:
It shows how the distance of an object changes over time. The slope of the graph gives the speed.
Speed-Time Graph:
It shows how the speed of an object changes over time. The area under the graph gives the distance covered.
2.5 Falling objects near the Earth have the same constant acceleration. Does this imply that a heavier object will fall faster than a lighter object?
No, heavier and lighter objects fall at the same rate near the Earth (ignoring air resistance) because the acceleration due to gravity is the same for all objects.
2.6 The vector quantities are sometimes written in scalar notation (not boldface). How is the direction indicated?
The direction of a vector quantity is indicated using an angle (e.g., \(30^\circ\)), a compass direction (e.g., north or east), or a unit vector.
2.7 A body is moving with uniform speed. Will its velocity be uniform? Give reason.
A body is moving with uniform speed. Will its velocity be uniform? Give reason.
No, the velocity will not be uniform if the direction of motion changes. Velocity depends on both speed and direction, so any change in direction results in a change in velocity.
2.8 Is it possible for a body to have acceleration? when moving with:(i) Constant velocity(ii) Constant speed.
(i) Constant Velocity:
No, a body with constant velocity has no acceleration because neither the speed nor the direction is changing.
(ii) Constant Speed:
Yes, a body can have acceleration if it is moving in a circular path, as its direction changes continuously.
Constructed Response Questions
2.1 Distance and displacement may or may not be equal in magnitude. Explain this statement.
(i)Distance:
It is the total length of the path traveled by an object, irrespective of direction.
(ii)Displacement:
It is the shortest straight-line distance between the initial and final positions, including direction.
If the object moves in a straight line without changing direction, distance and displacement will be equal. However, if the object moves along a curved or circular path, the distance will be greater than displacement.
2.2 When a bullet is fired, its velocity with which it leaves the barrel is called the muzzle velocity of the gun. The muzzle velocity of one gun with a longer barrel is less than that of another gun with a shorter barrel. In which gun is the acceleration of the bullet larger? Explain your answer.
Acceleration is the rate of change of velocity over time. In a longer barrel, the bullet has more time to accelerate due to the longer distance traveled inside the barrel. Therefore, the acceleration of the bullet is larger in the gun with a **longer barrel** because the force is applied over a greater distance and duration, allowing the bullet to accelerate more gradually.
2.3 For a complete trip, average velocity was calculated. Its value came out to be positive. Is it possible that its instantaneous velocity at any time during the trip had the negative value? Give justification of your answer.
Yes, it is possible.
Average velocity is calculated as the total displacement divided by the total time. A positive average velocity means the displacement is in the positive direction overall.
However, during the trip, the object could have temporarily moved in the opposite direction (negative direction), giving it a negative instantaneous velocity for that moment.
For example, a car moving back briefly before continuing forward still has a positive average velocity if the forward displacement is greater than the backward displacement.
Constructed Response Questions
2.4 A ball is thrown vertically upward with velocity \( v \). It returns to the ground in time \( T \). Which of the following graphs correctly represents the motion? Explain your reasoning.
The correct graph is (c).
Reason
When the ball is thrown upward, its velocity decreases linearly due to gravity until it reaches zero at the highest point (at \( T/2 \)).
Then, the ball starts to fall back, and its velocity increases linearly in the negative direction until it hits the ground.
The graph shows a straight line with a positive slope to zero (upward motion) and a straight line with a negative slope (downward motion).
2.5 The figure given below shows the distance-time graph for the travel of a cyclist. Find the velocities for the segments \( a \), \( b \), and \( c \).
To find velocity for each segment, use the formula
\[\text{Velocity} = \frac{\text{Distance}}{\text{Time}}\]
Segment \( a \)
\[\text{Distance covered} = 2.0 \, \text{km}, \, \text{Time taken} = 8 \, \text{min} = \frac{8}{60} \, \text{hr} = 0.133 \, \text{hr}\]
\[\text{Velocity} = \frac{2.0}{0.133} = 15 \, \text{km/hr}\]
Segment \( b \)
\[\text{Distance covered} = 0 \, \text{km}, \, \text{Time taken} = 4 \, \text{min} = \frac{4}{60} \, \text{hr} = 0.067 \, \text{hr}\]
\[\text{Velocity} = 0 \, \text{km/hr}\]
Segment \( c \)
\[\text{Distance covered} = -2.0 \, \text{km}, \, \text{Time taken} = 8 \, \text{min} = \frac{8}{60} \, \text{hr} = 0.133 \, \text{hr}\]
\[\text{Velocity} = \frac{-2.0}{0.133} = -15 \, \text{km/hr}\]
Final Velocities
\( a: 15 \, \text{km/hr} \)
\( b: 0 \, \text{km/hr} \)
\( c: -15 \, \text{km/hr} \)
2.6 Is it possible that the velocity of an object is zero at an instant of time, but its acceleration is not zero? If yes, give an example of such a case.
Yes, it is possible.
Example
When an object is thrown vertically upward, it reaches its highest point momentarily where the velocity becomes zero. However, the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)) is still acting on the object.
Comprehensive Questions
2.1 How can a vector be represented graphically? Explain.
A vector is represented graphically by a directed line segment.
Length of the Line Represents the magnitude of the vector.
Arrowhead Indicates the direction of the vector.
For example, a vector representing velocity might have an arrow pointing in the direction of motion, and its length represents the speed.
2.2 Differentiate between:(i) Rest and Motion || (ii) Speed and Velocity.
(i)Rest
An object is at rest when it does not change its position relative to a reference point. Example: A book on a table.
(i)Motion
An object is in motion when its position changes relative to a reference point. Example: A moving car.
(ii)Speed
The rate of change of distance. It is a scalar quantity. Example: 50 km/h.
(ii)Velocity
The rate of change of displacement. It is a vector quantity and has direction. Example: 50 km/h north.
2.3 Describe different types of motion. Also, give examples.
There are some types of the motion is given below.
Linear Motion
Motion along a straight line.
Example
A car moving on a straight road.
Rotational Motion
Motion around a fixed axis.
Example
Rotation of the Earth around its axis.
Periodic Motion
Motion that repeats at regular intervals.
Example
Swinging of a pendulum.
Random Motion
Motion that has no fixed path.
Example
Motion of a dust particle in the air.
2.4 Explain the difference between distance and displacement.
(i)Distance
The total path traveled by an object. It is a scalar quantity and always positive.
Example
A car traveling 10 km in a circle has a distance of 10 km.
(ii)Displacement
The shortest straight-line distance between the initial and final position of an object. It is a vector quantity and can be positive, negative, or zero.
Example
In the same circular journey, the displacement is 0 if the car returns to the starting point.
2.5 What do gradients of distance-time graph and speed-time graph represent? Explain it by drawing diagrams.
(i)Gradient of Distance-Time Graph
Represents speed. A steeper gradient indicates higher speed.
(ii)Gradient of Speed-Time Graph
Represents acceleration. A steeper gradient means greater acceleration.
Explanation
In a distance-time graph, the slope (\( \frac{\Delta s}{\Delta t} \)) gives the speed.
In a speed-time graph, the slope (\( \frac{\Delta v}{\Delta t} \)) gives the acceleration.
2.6 Prove that the area under a speed-time graph is equal to the distance covered by an object.
The area under a speed-time graph represents the distance covered because:
\[\text{Distance} = \text{Speed} \times \text{Time}\]
For example
For uniform speed (rectangle)
\[\text{Area} = \text{Speed} \times \text{Time} \]
For changing speed (triangle)
\[\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \]
This shows that the total area under the graph is equal to the distance traveled.
2.7 How equations of motion can be applied to the bodies moving under the action of gravity?
\( v = u + gt \)
\( s = ut + \frac{1}{2}gt^2 \)
\( v^2 = u^2 + 2gs \)
Here:
\( u \): Initial velocity
\( v \): Final velocity
\( g \): Acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \))
\( s \): Displacement
\( t \): Time
These equations help to analyze the motion of objects in free fall or those thrown upward.
Numerical Problems
2.1 Draw the representative lines of the following vectors:
(a) A velocity of 400 m s⁻¹ making an angle of 60° with x-axis.
(b) A force of 50 N making an angle of 120° with x-axis.
To Find
Draw the vectors as per the given angles.
Solution
(a) Draw a line with a magnitude of 400 m/s at an angle of 60° above the x-axis.
(b) Draw a line with a magnitude of 50 N at an angle of 120° (above the negative x-axis).
Since this involves drawing, I recommend plotting this using a diagramming tool or graph paper.
2.2 A car is moving with an average speed of 72 km h⁻¹. How much time will it take to cover a distance of 360 km?
To Find
Time taken to cover the distance.
Solution
Given
Speed, \( v = 72 \, \text{km/h} \)
Distance, \( d = 360 \, \text{km} \)
Formula:
\[t = \frac{d}{v}\]
Substitute values
\[t = \frac{360}{72} = 5 \, \text{hours}\]
Answer : \( t = 5 \, \text{h} \)
2.3 A truck starts from rest. It reaches a velocity of 90 km h⁻¹ in 50 seconds. Find its average acceleration.
To Find
Average acceleration.
Solution
Given
Initial velocity, \( u = 0 \, \text{m/s} \)
Final velocity, \( v = 90 \, \text{km/h} = 25 \, \text{m/s} \) (converted to m/s)
Time, \( t = 50 \, \text{s} \)
Formula
\[a = \frac{v – u}{t}\]
Substitute values
\[a = \frac{25 – 0}{50} = 0.5 \, \text{m/s²}\]
Answer \( a = 0.5 \, \text{m/s²} \)
2.4 A car passes a green traffic signal while moving with a velocity of 5 m s⁻¹. It then accelerates to 1.5 m s⁻². What is the velocity of the car after 5 seconds?
To Find
Final velocity of the car.
Solution
Given
Initial velocity, \( u = 5 \, \text{m/s} \)
Acceleration, \( a = 1.5 \, \text{m/s²} \)
Time, \( t = 5 \, \text{s} \)
Formula
\[v = u + at\]
Substitute values
\[v = 5 + (1.5 \times 5) = 5 + 7.5 = 12.5 \, \text{m/s}\]
Answer \( v = 12.5 \, \text{m/s} \)
2.5 A motorcycle initially travelling at 18 km h⁻¹ accelerates at constant rate of 2 m s⁻². How far will the motorcycle go in 10 seconds?
To Find
Distance traveled.
Solution
Given
Initial velocity, \( u = 18 \, \text{km/h} = 5 \, \text{m/s} \) (converted to m/s)
Acceleration, \( a = 2 \, \text{m/s²} \)
Time, \( t = 10 \, \text{s} \)
Formula
\[s = ut + \frac{1}{2}at^2\]
Substitute values
\[s = (5 \times 10) + \frac{1}{2}(2 \times 10^2)
s = 50 + 100 = 150 \, \text{m}\]
Answer \( s = 150 \, \text{m} \)
2.6 A wagon is moving on the road with a velocity of 54 km h⁻¹. Brakes are applied suddenly. The wagon covers a distance of 25 m before stopping. Determine the acceleration of the wagon.
To Find
Acceleration of the wagon.
Solution
Given
Initial velocity, \( u = 54 \, \text{km/h} = 15 \, \text{m/s} \) (converted to m/s)
Final velocity, \( v = 0 \, \text{m/s} \)
Distance, \( s = 25 \, \text{m} \)
Formula
\[v^2 = u^2 + 2as\]
Rearranging for \( a \)
\[a = \frac{v^2 – u^2}{2s}\]
Substitute values
\[a = \frac{0 – 15^2}{2 \times 25} = \frac{-225}{50} = -4.5 \, \text{m/s²}\]
Answer: \( a = -4.5 \, \text{m/s²} \)
2.7 A stone is dropped from a height of 45 m. How long will it take to reach the ground? What will be its velocity just before hitting the ground?
To Find
(a) Time to hit the ground.
(b) Final velocity.
Solution
Given
Height, \( h = 45 \, \text{m} \)
Acceleration due to gravity, \( g = 9.8 \, \text{m/s²} \)
Initial velocity, \( u = 0 \, \text{m/s} \)
(a) Time
Formula
\[h = \frac{1}{2}gt^2\]
Rearranging for \( t \)
\[t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 45}{9.8}} = \sqrt{9.18} \approx 3 \, \text{s}\]
(b) Final Velocity:
Formula:
\[v = \sqrt{2gh}\]
Substitute values
\[v = \sqrt{2 \times 9.8 \times 45} = \sqrt{882} \approx 30 \, \text{m/s}\]
Answer: \( t = 3 \, \text{s}, v = 30 \, \text{m/s} \)
2.8 A car travels 10 km with an average velocity of 20 m/s. Then it travels in the same direction through a diversion at an average velocity of 4 m/s for the next 0.8 km. Determine the average velocity of the car for the total journey.
To Find
Average velocity of the car for the entire journey.
Solution
Given
Distance \( d_1 = 10 \, \text{km} = 10,000 \, \text{m} \)
Velocity \( v_1 = 20 \, \text{m/s} \)
Distance \( d_2 = 0.8 \, \text{km} = 800 \, \text{m} \)
Velocity \( v_2 = 4 \, \text{m/s} \)
1. Find the total distance
\[\text{Total Distance}, d_{\text{total}} = d_1 + d_2 = 10,000 + 800 = 10,800 \, \text{m}\]
2. Find the time taken for each segment
\[t_1 = \frac{d_1}{v_1} = \frac{10,000}{20} = 500 \, \text{s}\]
\[t_2 = \frac{d_2}{v_2} = \frac{800}{4} = 200 \, \text{s}\]
3. Find the total time taken
\[t_{\text{total}} = t_1 + t_2 = 500 + 200 = 700 \, \text{s}\]
4. Find the average velocity
\[v_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{10,800}{700} = 15.4 \, \text{m/s}\]
Answer: \( v_{\text{avg}} = 15.4 \, \text{m/s} \)
2.9 A ball is dropped from the top of a tower. The ball reaches the ground in 5 seconds. Find the height of the tower and the velocity of the ball with which it strikes the ground.
To Find
(a) Height of the tower.
(b) Final velocity of the ball.
Solution
Given
Time \( t = 5 \, \text{s} \)
Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \)
Initial velocity \( u = 0 \, \text{m/s} \)
1. Find the height of the tower
Use the equation of motion
\[s = ut + \frac{1}{2}gt^2\]
\[s = 0 \times 5 + \frac{1}{2} \times 9.8 \times 5^2\]
\[s = \frac{1}{2} \times 9.8 \times 25 = 122.5 \, \text{m}\]
2. Find the final velocity
Use the equation of motion:
\[v = u + gt\]
\[v = 0 + 9.8 \times 5 = 49 \, \text{m/s}\]
Answer
Height = \( 122.5 \, \text{m} \)
Final Velocity = \( 49 \, \text{m/s} \)
(Note: Rounded as \( 125 \, \text{m} \) and \( 50 \, \text{m/s} \) in the question.)
2.10 A cricket ball is hit so that it travels straight up in the air. An observer notes that it took 3 seconds to reach the highest point. What was the initial velocity of the ball? If the ball was hit 1 m above the ground, how high did it rise from the ground?
To Find
(a) Initial velocity of the ball.
(b) Maximum height reached by the ball.
Solution
Given:
Time to reach the highest point \( t = 3 \, \text{s} \)
Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \)
Height above the ground where the ball was hit \( h_{\text{initial}} = 1 \, \text{m} \)
1. Find the initial velocity
At the highest point, the velocity becomes zero (\( v = 0 \)). Use the equation:
\[v = u – gt\]
Rearranging for \( u \):
\[u = v + gt = 0 + 9.8 \times 3 = 29.4 \, \text{m/s}\]
2.Find the maximum height from the point of release
Use the equation
\[v^2 = u^2 – 2gh\]
Rearranging for \( h \)
\[h = \frac{u^2 – v^2}{2g} = \frac{(29.4)^2 – 0}{2 \times 9.8} = \frac{864.36}{19.6} = 44.1 \, \text{m}\]
3.Find the total height from the ground
\[h_{\text{total}} = h_{\text{initial}} + h = 1 + 44.1 = 45.1 \, \text{m}\]
Answer
Initial Velocity = \( 29.4 \, \text{m/s} \) (rounded to \( 30 \, \text{m/s} \))
Maximum Height = \( 45.1 \, \text{m} \) (rounded to \( 46 \, \text{m} \))
