ICSE Solutions for Chapter 2 Motion in One Dimension Class 9 Physics Selina Publishers
Exercise -2(A)
1. Differentiate between the scalar and vector quantities, giving two examples of each.
Solution
Scalar |
Vector |
They are expressed only by their magnitudes. |
They are expressed by magnitude as well as direction. |
They can be added, subtracted, multiplied or divided by simple arithmetic methods. |
They can be added, subtracted or multiplied following a different algebra. |
They are symbolically written by English letter. |
They are symbolically written by their English letter with an arrow on top of the letter. |
Example: mass, speed |
Example: force, velocity |
2. State whether the following quantity is a scalar or vector?
(a) Pressure
(b) Force
(c) Momentum
(d) Energy
(e) Weight
(f) Speed
Solution
a) Pressure is a scalar quantity.
b) Momentum is a vector quantity.
c) Weight is a vector quantity.
d) Force is a vector quantity.
e) Energy is a scalar quantity.
f) Speed is a scalar quantity.
3. When is a body said to be at rest?
Solution
A body is said to be at rest if it does not change its position with respect to its immediate surroundings.
4. When is a body said to be in motion?
Solution
A body is said to be in motion if it changes its position with respect to its immediate surroundings.
5. What do you mean by motion in one direction?
Solution
When a body moves along a straight line path, its motion is said to be one-dimensional motion.
6. Define displacement. State its unit.
Solution
The shortest distance from the initial to the final position of the body is called the magnitude of displacement. It is in the direction from the initial position to the final position.
Its SI unit is metre (m).
7. Differentiate between distance and displacement.
Solution
Distance |
Displacement |
It is a scalar quantity. |
It is a vector quantity. |
The distance is the length of path travelled by the body. |
The displacement is the shortest length in direction from initial to the final position. |
It is always positive. |
It can be positive or negative depending on the direction of the body. |
Can be more than or equal to the magnitude of displacement. |
Can be less than or equal to the distance, but never greater than the distance. |
8. Can displacement be zero even if distance is not zero? Give one example to explain your answer.
Solution
Yes, displacement can be zero even if the distance is not zero.
For example, when a body is thrown vertically upwards from a point A on the ground, after sometime it comes back to the same point A. Then, the displacement is zero, but the distance travelled by the body is not zero (it is 2h; h is the maximum height attained by the body).
9. When is the magnitude of displacement equal to the distance?
Solution
The magnitude of displacement is equal to the distance when the motion is in a fixed direction.
10. Define velocity. State its unit.
Solution
The velocity of a body is the distance travelled per second by the body in a specified direction.
Its SI unit is metre/second (m/s).
11. Define speed. What is its S.I. unit?
Solution
The speed of a body is the rate of change of distance with time.
Its SI unit is metre/second (m/s).
12. Distinguish between speed and velocity.
Solution
Speed |
Velocity |
Scalar quantity |
Vector quantity |
Rate of change of distance with time |
Rate of change of displacement of a body with time |
Speed is always positive-it is the magnitude of velocity. |
The velocity is given a positive or negative sign depending upon its direction of motion. |
Average speed is never zero. |
Average velocity can be zero |
When the body returns to its initial position, speed will not be zero. |
When the body returns to its initial position, velocity can be zero. |
13. Which of the quantity speed or velocity gives the direction of motion of body.
Solution
Velocity gives the direction of motion of the body.
14. When is the instantaneous speed same as the average speed?
Solution
Instantaneous velocity is equal to average velocity if the body is in uniform motion.
15. Distinguish between uniform velocity and variable velocity.
Solution
Uniform velocity |
Variable velocity |
If a body travels equal distances in equal intervals of time along a particular direction, then the body is said to be moving with a uniform velocity. |
iIf a body travels unequal distances in a particular direction in equal intervals of time or it moves equal distances in equal intervals of time but its direction of motion does not remain same, then the velocity of the body is said to be variable (or non-uniform). |
16. Distinguish between average speed and average velocity.
Solution
Average speed |
Average velocity |
It is the ratio of the total distance travelled by the body to the total time of journey. |
If the velocity of a body moving in a particular direction changes with time, then the ratio of displacement to the time taken in entire journey is called its average velocity. |
It is never zero. |
It can be zero even if its average speed is not zero. |
17. Give an example of motion of a body moving with a constant speed, but with a variable velocity. Draw a diagram to represent such a motion.
Solution
The motion of a body in a circular path with uniform speed has a variable velocity because in the circular path, the direction of motion of the body continuously changes with time.
18. Give an example of motion in which average speed is not zero, but average velocity is zero.
Solution
If a body starts its motion from a point and comes back to the same point after a certain time, then the displacement is zero, average velocity is also zero, but the total distance travelled is not zero, and therefore, the average speed in not zero.
19. Define acceleration. State its S.I unit.
Solution
Acceleration can be defined as the rate of change of velocity with time. It can also be defined as the increase in velocity per second. The S.I. unit of acceleration is meter per second square or ms^{-2}.
20. Distinguish between acceleration and retardation.
Solution
Acceleration |
Retardation |
It is the increase in velocity per second. |
It is the decrease in the velocity per second. |
If the final velocity is greater than initial velocity, acceleration is positive. |
If the final velocity is lesser than initial velocity, acceleration becomes negative, hence it is the retardation. |
21. Differentiate between uniform acceleration and variable acceleration.
Solution
Uniform acceleration |
Variable acceleration |
Acceleration is said to be uniform when equal changes in velocity take place in equal intervals of time. |
If the change in velocity is not the same in the same intervals of time, the acceleration is said to be variable. |
Example — Motion of a body under gravity(free fall of a body) |
Example — Motion of a vehicle on a hilly road |
22. What is meant by the term retardation? Name its S.I. unit.
Solution
Retardation is the decrease in velocity per second.
Its SI unit is metre/second^{2} (m/s^{2}).
23. Which of the quantity, velocity or acceleration determines the direction of motion?
Solution
Velocity determines the direction of motion.
24. Give one example of each type of the following motion:
(a) Uniform velocity
(b) Variable velocity
(c) Variable acceleration
(d) Uniform retardation
Solution
(a) Example of uniform velocity: A body, once started, moves on a frictionless surface with uniform velocity.
(b) Example of variable velocity: A ball dropped from some height is an example of variable velocity.
(c) Example of variable acceleration: The motion of a vehicle on a crowded road is with variable acceleration.
(d) Example of uniform retardation: If a car moving with a velocity ‘v’ is brought to rest by applying brakes, then such a motion is an example of uniform retardation.
25. The diagram below shows the pattern of the oil on the road, dripping at a constant rate from a moving car. What information do you get from it about the motion of car?
Solution
Initially as the drops are equidistant, we can say that the car is moving with a constant speed but later as the distance between the drops starts decreasing, we can say that the car slows down.
26. Define the term acceleration due to gravity. State its average value.
Solution
When a body falls freely under gravity, the acceleration produced in the body due to the Earth’s gravitational acceleration is called the acceleration due to gravity (g). The average value of g is 9.8 m/s^{2}.
27. ‘The value of g remains same at all places on the earth surface’, Is this statement true? Give reason for your answer.
Solution
No. The value of ‘g’ varies from place to place. It is maximum at poles and minimum at the Equator on the surface of the Earth.
28. If a stone and a pencil are dropped simultaneously in vaccum from the top of a tower, which of the two will reach the ground first? Give reason.
Solution
In vacuum, both will reach the ground simultaneously because acceleration due to gravity is same (=g) on both objects.
Multiple choice type
1. A vector quantity is:
(a) Work
(b) Pressure
(c) Distance
(d) Velocity
Solution
(d) Velocity
A vector quantity has both magnitude and direction.
2. The S.I. unit of velocity is:
(a) Kmh^{-1}
(b) m min^{-1}
(c) km min^{-1}
(d) ms^{-1}
Solution
(d) ms^{-2}
Velocity is the rate of change of displacement (m) of a body with time(s).
3. The unit of retardation is:
(a) ms^{-1}
(b) ms^{-2}
(c) m
(d) ms
Solution
(b) ms^{-2}
Retardation is negative value of acceleration.
4. A body when projected up with an initial velocity u goes to a maximum height h in time t and then comes back at the point of projection. The correct statement is:
(a) The average velocity is 2h/t
(b) The acceleration is zero
(c) The final velocity on reaching the point of projection is 2u
(d) The displacement is zero
Solution
(d) The displacement is zero
The displacement is zero if the initial and the final position of the body is same.
5. 18km h’' is equal to:
(a) 10ms^{-1}
(b) 5 ms^{-1}
(c) 18 ms^{-1}
(d) 1.8 ms^{-1}
Solution
(b) 5 ms^{-1}
18 kmh^{-1} = (18×1000)/(60×60) = 5 ms^{-1}
Numericals
1. The speed of a car is 72km h^{-1}. Express it in ms^{-1}.
Solution
Speed of car = 72kmh^{-1}
2. Express 15 ms^{-1} in km h^{-1}
Solution
3. Express each of the following in ms^{-1}:
(a) 1 km h^{-1}
(b) 18 km min^{-1}
Solution
(a) Expressing 1km h^{-1} in ms^{-1}
1km = 1000 m
1h = 3600 s
(b) Expressing 18km min^{-1} in ms^{-1}
1km = 1000 m
1min = 60 s
4. Arrange the following speeds in increasing order:
10 ms^{-1}, 1 km min^{-1}, 18 km h^{-1}
Solution
Expressing all the three in ms^{-1}
10 ms^{-1} is already in ms^{-1}
1 km min^{-1} = 1 × 1000/60 = 16.67 ms^{-1}
18 kmh^{-1} = (18×1000)m/(60×60)s = 5 ms^{-1}
Hence, the increasing order is 18 km h^{-1}, 10 ms^{-1}, 1 km min^{-1}
5. A train takes 3h to travel from Agra to Delhi with a uniform speed of 65 km h^{-1}. Find the distance between the two cities.
Solution
Total time taken = 3 hours
Speed of the train = 65 km/hr
Distance travelled = speed × time =65×3= 195 km
6. A car travels first 30km with a uniform speed of 60km h^{-1} and then next 30km with a uniform speed of 40 km h^{-1}. Calculate: (i) the total time of journey, (ii) the average speed of the car.
Solution
For the first 30 km travelled, speed = 60 km/h.
Thus time taken (t_{1}) = Distance/speed
= (30/60) h^{-1}
=0.5h^{-1} or 30 min.
For the next 30 km travelled, speed = 40 km/h
Thus time taken (t_{2}) = Distance/speed
= (30/40) h^{-1}
=0.75 h^{-1} or 45 min.
(i) Total time = (30 + 45) min
= 75 min or 1.25h.
(ii) Average speed of the car = Total distance travelled/total time taken
= 60km/1.25hr = 48 km/h
7. A train takes 2h to reach station B from station A, and then 3 h to return from station B to station A. The distance between the two stations is 200km. Find:
(i) The average speed
(ii) The average velocity of the train
Solution
Here, total distance = (200 + 200) km = 400 km
Total time taken = (2+3) h=5h
(i) Average speed = Total distance travelled/total time taken
= 400km/5h = 80 km/h
(ii) Average velocity of the train is zero because the train stops at the same point from where it starts, i.e. the displacement is zero.
8. A car moving on a straight path covers a distance of 1km due east in 100s. What is (i) the speed and (ii) the velocity of car?
Solution
(i) Speed of the car = Distance/time taken
(ii) Velocity of car = Speed with direction = 10 m/s due east
9. A body starts from rest and acquires a velocity 10 ms^{-1} in 2s. Find its acceleration.
Solution
Here, final velocity = 10 m/s
Initial velocity = 0 m/s
Time taken = 2s
Acceleration = (Final Velocity - Initial Velocity)/time
= (10/2) m/s^{2}
= 5 m/s^{2}
10. A car starting from rest acquires a velocity 180 ms^{-1} in 0.05h. Find the acceleration.
Solution
Here, final velocity = 180 m/s
Initial velocity = 0 m/s
Time taken = 0.05 h or 180s
Acceleration = (Final Velocity - Initial Velocity)/time
= (180-0)/180 m/s^{2}
= 1 m/s^{2}
11. A body is moving vertically upwards. Its velocity changes at a constant rate from 50 ms^{-1} to 20 ms^{-1} in 3s. What is its acceleration?
Solution
Here, final velocity = 20 m/s
Initial velocity = 50 m/s
Time taken = 3 s
Acceleration = (Final Velocity - Initial Velocity)/time
= (20 - 50)/3 m/s^{2}
= -10 m/s^{2}
Negative sign here indicates that the velocity decreases with time, so retardation is 10 m/s.
12. A toy car initially moving with a uniform velocity of 18km h^{-1} comes to a stop in 2s. Find the retardation of the car in S.I. units.
Solution
Here, final velocity = 18 km/h or 5 m/s
Initial velocity = 0 km/h
Time taken = 2s
Acceleration = (Final Velocity - Initial Velocity)/time
= (5-0)/2 m/s^{2}
= 2.5 m/s^{2}
13. A car accelerates at a rate of 5 ms^{-2}. Find the increase in its velocity in 2s.
Solution
Acceleration = Increase in velocity/time taken
Therefore, increase in velocity = Acceleration × time taken
(5×2) m/s =10 m/s
14. A car is moving with a velocity 20m/s. The brakes are applied to retard it at a rate of 2m/s^{2}. What will be the velocity after 5s of applying the brakes?
Solution
Initial velocity of the car, u = 20 m/s
Retardation = 2 m/s^{2}
Given time, t=5s
Let ‘v’ be the final velocity.
We know that, Acceleration = Rate of change of velocity /time
= (Final velocity - Initial velocity)/time
Or, -2 = (v-20)/5
Or, -10 = v = 20
Or, v = -20 + 10 m/s
Or, v = -10 m/s
Negative sign indicates that the velocity is decreasing.
15. A bicycle initially moving with a velocity 5m/s accelerates for 5s at a rate of 2 m/s^{2}. What will be its final velocity?
Solution
Initial velocity of the bicycle, u = 5 m/s
Acceleration = 2 m/s^{2}
Given time, t=5s
Let ‘v’ be the final velocity.
We Know that, acceleration = Rate of change of velocity/time
= (Final velocity — Initial velocity)/time
Or, 2 = (v= 5)/5
Or, 10 = (v - 5)
Or, v=-5-10
Or, v = -15 m/s
Negative sign indicates that the velocity is decreasing.
16. A car is moving in a straight line with speed 18km h"'. It is stopped in 5s by applying the brakes. Find: (i) the speed of car in m/s, (ii) the retardation and (iii) the speed of car after 2s of applying the brakes.
Solution
Initial velocity of the bicycle, u = 18 km/hr
Time taken, t=55
Final velocity, v = 0 m/s (As the car comes to rest)
(i) Speed in m/s = (18×1000)/(1×3600) = 5 m/s
(ii) Retardation = (Final velocity - Initial velocity)/time taken
Or, Retardation = (0-5)/5 ms^{-2} = 1 ms^{-2}
(iii) Let ‘V’ be the speed of the car after 2 s of applying the brakes.
Then, Acceleration = (V - 5)/ 2
Or, -1 = (V-5)/2
Or, V=-2+5
Or, V = 3 m/s^{2}
Exercise 2(B)
1. For the motion with uniform velocity, how is the distance travelled related to the time?
Solution
Distance is directly proportional to time.
2. What information about the motion of a body are obtained from the displacement-time graph?
Solution
From displacement-time graph, the nature of motion (or state of rest) can be understood. The slope of this graph gives the value of velocity of the body at any instant of time, using which the velocity-time graph can also be drawn.
3. (a) What does the slope of a displacement-time graph represent?
(b) Can displacement-time sketch be parallel to the displacement axis? Give reason to your answer.
Solution
(a) Slope of a displacement-time graph represents velocity.
(b) The displacement-time graph can never be parallel to the displacement axis because such a line would mean that the distance covered by the body in a certain direction increases without any increase in time, which is not possible.
4. Draw a displacement-time graph for a boy going to school with a uniform velocity.
Solution
5. State how the velocity-time graph can be used to find (i) the acceleration of a body, (ii) the distance travelled by the body in a given time, and (iii) the displacement of the body in a given time.
Solution
(i) The slope of the velocity-time graph gives the value of acceleration.
(ii) The total distance travelled by a body in a given time is given by the area enclosed between the velocity-time graph and X-axis (without any sign).
(iii) The displacement of a body in a given time is given by the area enclosed between the velocity-time graph and X-axis (with proper signs).
6. What can you say about the nature of motion of a body if its displacement-time graph is
(a) A straight line parallel to time axis?
(b) A straight line inclined to the time axis with an acute angle?
(c) A straight line inclined to the time axis with an obtuse angle?
(d) A curve
Solution
(a) There is no motion, the body is at rest.
(b) It depicts that the body is moving away from the starting point with uniform velocity.
(c) It depicts that the body is moving towards the starting point with uniform velocity.
(d) It depicts that the body is moving with variable velocity.
7. The figure shows displacement-time graph of two vehicles A and B along a straight road. Which vehicle is moving faster? Give reason.
Solution
Vehicle A is moving with a faster speed because the slope of line A is more than the slope of line B.
8. State the type of motion represented by the following sketches in the figure. Give example of each type of motion.
Solution
(a) The graph depicts uniform accelerated motion.
Example - It is the motion of a body that is liberated downwards.
(b) The graph indicates the motion of a body with variable retardation.
Example — a car reaching its destination.
9. Draw a velocity-time graph for a body moving with an initial velocity u and uniform acceleration a. Use this graph to find the distance travelled by the body in time t.
Solution
In this graph, initial velocity = u
Velocity at time t = v
Let acceleration be ‘a’
Time =t
Then, distance travelled by the body in ‘t’ s = area between the v-t graph and X-axis
Or, distance travelled by the body in ‘t’ s = area of the trapezium OABD
= (1/2) × (sum of parallel sides) × (perpendicular distance between them)
= (1/2) × (u+v) × (t)
=(u+v)t/2
10. What does the slope of velocity-time graph represent?
Solution
The slope of the velocity-time graph represents acceleration.
11. Figure shows the velocity-time graph for two cars A and B moving in same direction. Which car has the greater acceleration? Give reason to your answer.
Solution
Car B has greater acceleration because the slope of line B is more than the slope of line A.
12. Draw the shape of the velocity-time graph for a body moving with (a) uniform velocity, (b) uniform acceleration.
Solution
Velocity-time for a body moving with uniform velocity and uniform acceleration.
13. The velocity-time graph for a uniformly retarded body is a straight line inclined to the time axis with an obtuse angle. How is retardation calculated from the velocity-time graph?
Solution
Retardation is calculated by finding the negative slope.
14. Figure shows the displacement-time graph for four bodies A, B, C and D. In each case state what information do you get about the acceleration (zero, positive or negative).
Solution
For body A: The graph is a straight line. So, the slope gives constant velocity. Hence, the acceleration for body A is zero.
For body B: The graph is a straight line. So, the slope gives constant velocity. Hence, the acceleration for body B is also zero.
For body C: The slope of the graph is decreasing with time. Hence, the acceleration is negative.
For body D: The slope of the graph is increasing with time. Hence, the acceleration is positive.
15. Draw a graph for acceleration against time for a uniformly accelerated motion. How can it be used to find the change in speed in a certain interval of time?
Solution
Changing speed at a particular interval of time can be obtained by the area enclosed between the straight line and the time axis for that interval of time.
16. Draw a velocity-time graph for the free fall of a body under gravity, starting from rest. Take g=10ms^{2}
Solution
17. How is the distance related with time for the motion under uniform acceleration such as the motion of a freely falling body?
Solution
For motion under uniform acceleration, such as the motion of a freely falling body, distance is directly proportion to the square of the time.
18. A body falls freely from a certain height. Show graphically the relation between the distance fallen and square of time. How will you determine g from this graph?
Solution
The value of acceleration due to gravity (g) can be obtained by doubling the slope of the S – t^{2} graph for a freely falling body.
Multiple choice type
1. The velocity-time graph of a body in motion is a straight line inclined to the time axis. The correct statement is:
(a) Velocity is uniform
(b) Acceleration is uniform
(c) Both velocity and acceleration are uniform
(d) Neither velocity nor acceleration is uniform
Solution
(b) Acceleration is uniform
If the velocity-time graph is a straight line inclined to the time axis, the motion is with uniform acceleration.
2. For a uniformly retarded motion, the velocity-time graph is:
(a) A curve
(b) A straight line parallel to the time axis
(c) A straight line perpendicular to the time axis
(d) A straight line inclined to the time axis
Solution
(d) A straight line inclined to the time axis
3. For the uniform motion:
(a) The distance-time graph is a straight line parallel to the time axis
(b) The speed-time graph is a straight line inclined to the time axis
(c) The speed-time graph is a straight line parallel to the time axis
(d) The acceleration-time graph is a straight line parallel to the time axis.
Solution
(c) The speed-time graph is a straight line parallel to the time axis
Numericals
1. Figure shows the displacement-time graph for the motion of a body. Use it to calculate the velocity of body at t=1s, 2s and 3s, then draw the velocity-time graph for it in figure (b).
Solution
Velocity of body at t = 1s is 2 m/s
Velocity of body at t = 2s is 4 m/s
Velocity of body at t = 3s is 6 m/s
2. Following table gives the displacement of a car at different instants of time.
Time (s) |
0 |
1 |
2 |
3 |
4 |
Displacement (m) |
0 |
5 |
10 |
15 |
20 |
(a) Draw the displacement-time sketch and find the average velocity of car.
(b) What will be the displacement of car at (i) 2.5s and (ii) 4.5s?
Solution
Displacement-time graph
From the part AB of the graph,
Average velocity = (Displacement at B - Displacement at A)/Time taken
= (30 - 20) m/(6-4)s
= (10/2) m/s
= 5m/s
(b) (i) From the graph, the displacement of car at 2.5 s is 12.5 m.
(ii) From the graph, the displacement of car at 4.5 s is 22.5 m.
3. A body is moving in a straight line and its displacement at various instants of time is given in the following table:
Time (s) |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
Displacement (m) |
2 |
6 |
12 |
12 |
12 |
18 |
22 |
24 |
Plot displacement-time graph and calculate:
(i) Total distance travelled in interval 1s to 5s,
(ii) Average velocity in time interval 1s to 5s.
Solution
(i) Total distance travelled in interval 1s to 5s = 18m-6m= 12m.
(ii) Average velocity = Total displacement in the given time interval/Time interval, I.e. 1s to 5s,
Or, Average velocity = 12m/4s = 3 m/s.
4. Figure shows the displacement of a body at different times.
(a) Calculate the velocity of the body as it moves for time interval (i) 0 to 5s, (ii) 5s to 7s and
(iii) 7s to 9s.
(b) Calculate the average velocity during the time interval 5s to 9s.
[Hint: From 5s to 9s, displacement=7m -3m = 4m]
Solution
(a) (i) Velocity from 0 to 5 s = Displacement /time
= (3/5) m/s
= 0.6 m/s
(ii) Velocity from 5s to 7s = Displacement /time
= (0/2) m/s
= 0m/s.
(iii) Velocity from 7s to 9s = Displacement /time
= (7 - 3)/(9 - 7) m/s
= (4/2) m/s
= 2m/s
(b) From, 5s to 9s, displacement = 7m - 3m = 4m.
Time elapsed between 5s to 9s=45
Average velocity = Displacement/time
= (4/4) m/s
= 1m/s
5. From the displacement-time graph of a cyclist, given in figure, find:
(i) The average velocity in the first 4s,
(ii) The displacement from the initial position at the end of 10s
(iii) The time after which he reaches the starting point.
Solution
(i) Displacement in first 4s = 10 m
Therefore, the average velocity = Displacement/time
= (10/4) m/s = 2.5 m/s
(ii) Initial position = 0 m
Final position at the end of 10 s = -10m
Displacement = Final position - Initial position
= (-10)m – 0 = -10m
6. Figure ahead represents the displacement-time sketch of motion of two cars A and B. Find:
(i) The distance by which the car B was initially ahead of car A.
(ii) The velocities of car A and car B
(iii) The time in which car A catches car B
(iv) The distance from start when the car A will catch the car B
Solution
(i) As per the graph, the car B was ahead of car A by 40km
(ii) The lines of car A and B as per the graph are straight indicating they have uniform velocities.
Displacement of car A:
Time = 1 hour, distance = 40km
Velocity = displacement/time
= 40/1 = 40km/h
Displacement of car B:
Time = 4 hour, distance = (120-40) = 80km
Velocity = displacement/time
= 80/4 = 20km/h
(iii) As per the graph, both cars A and B intersect at a point, it is at this point the car A catches car B, i.e., 2hours
(iv) As observed in the graph, the distance from the start when car A will catch car B is 80km.
7. A body at rest is made to fall from the top of a tower. Its displacement at different instants is given in the following table:
Time (s) |
0.1 |
0.2 |
0.3 |
0.4 |
0.5 |
0.6 |
Displacement (m) |
0.05 |
0.20 |
0.45 |
0.80 |
1.25 |
1.80 |
Draw a displacement-time graph and state whether the motion is uniform or non-uniform?
Solution
The motion is non-uniform as the displacement-time graph is a curve.
8. Figure shows the velocity-time graph for the motion of a body. Use it to find the displacement of the body at t=1s, 2s, 3s and 4s, then draw the displacement-time graph for it on figure (b).
Solution
As per the graph,
Velocity(m/s) |
Time(s) |
Displacement = velocity×time |
1 |
1 |
1 |
2 |
2 |
4 |
3 |
3 |
9 |
9. Figure below shows a velocity-time graph for a car starting from rest. The graph has three parts AB, BC and CD.
(i) State how is the distance travelled in any part determined from this graph
(ii) Compare the distance travelled in part BC with the distance travelled in part AB.
(iii) Which part of graph shows motion with uniform (a) Velocity (b) acceleration (c) retardation?
(iv) (a) Is the magnitude of acceleration higher or lower than that of retardation? Give a reason,
(b) Compare the magnitude of acceleration and retardation.
Solution
(i) Distance travelled in any part of the graph can be determined by finding the area enclosed by the graph in that part with the time axis.
(ii) Distance travelled in part BC = Area of the rectangle tBC 2t = base × height.
= (2t-t) × v_{o}
= v_{o}t
Distance travelled in part AB = Area of the triangle ABt
= (1/2) × base × height
= (1/2) × t × v_{o}
= (1/2) v_{o}t
Therefore, distance travelled in part BC: distance travelled in part AB = 2:1.
(iii) (a) BC shows motion with uniform velocity.
(b) AB shows motion with uniform acceleration.
(c) CD shows motion with uniform retardation.
(iv) (a) The magnitude of acceleration is lower as the slope of line AB is less than that of line CD.
(b) Slope of line AB = v_{o}/t
Slope of line CD = v_{o}/0.5t
Slope of line AB/Slope of line CD = (v_{o}/t)/(v_{o}/0.5t)
Slope of line AB: Slope of line CD :: 1:2.
10. The velocity-time graph of a moving body is given below in figure
Find:
(i) The acceleration in parts AB, BC and CD.
(ii) Displacement in each part AB, BC, CD, and
(iii) Total displacement.
Solution
(i) Acceleration in the part AB = Slope of AB = tan (∠BAD)
= (30/4) ms^{-2}
= 7.5 ms^{-2}
Acceleration in the part BC = 0 ms^{-2}
Acceleration in the part CD = slope of CD = -tan (∠CDA)
= -(30/2) ms^{-2}
= -15 ms^{-2}
(ii) Displacement of part AB = Area of △AB4 = (1/2) (4) (30)
= 60m
Displacement of part BC = Area of rectangle 4BC8
= (30) × (4) = 120m
Displacement of part CD = Area of △C8D = (1/2) (2) (30)
= 30m
(iii) Total displacement = Displacement of part AB + Displacement of part BC + Displacement of part CD
= 60 + 120 + 30 = 210m
11. A ball moves on a smooth floor in a straight line with a uniform velocity 10m/s for 6s. At t=6s, the ball hits a wall and comes back along the same line to the starting point with same speed. Draw the velocity-time graph and use it to find the total distance travelled by the ball and its displacement.
Solution
Distance travelled in first 6 s = velocity × time
=10m/s ×6
= 60 m/s
Distance travelled in next 6 s = velocity × time
= 10m/s × 6
= 60 m/s
Total distance travelled in 12 s = (60 + 60) m= 120m
Total displacement = 0, as the ball returns its starting point.
12. Figure shows the velocity-time graph of a particle moving in a straight line.
(i) State the nature of motion of particle.
(ii) Find the displacement of particle at t=6s.
(iii) Does the particle change its direction of motion?
(iv) Compare the distance travelled by the particle from 0 to 4s and from 4s to 6s.
(v) Find the acceleration from 0 to 4s and retardation from 4s to 6s.
Solution
(i) From 0 to 4 seconds, the motion is uniformly accelerated and from 4 to 6 seconds, the motion is uniformly retarded.
(ii) Displacement of the particle at 6 s = (1/2) (6) (2) =6m
(iii) The particle does not change its direction of motion.
(iv) Distance travelled by the particle from 0 to 4s (D1) = (1/2) (4) (2) =4m
Distance travelled by the particle from 4 to 6s (D2) = (1/2) (2) (2) = 2m
D1:D2:: 4:2
D1:D2:: 2:1
(v) Acceleration from 0 to 4 s = (2/4) ms^{-2} or 0.5 ms^{-2}
Retardation from 4 s to 6 s = (2/2) ms^{-2} or 1 ms^{-2}.
Exercise 2(C)
1. Write three equations of uniformly accelerated motion relating the initial velocity (u), final velocity (v), time (t), acceleration (a) and displacement (S).
Solution
The three equations of uniformly accelerated motion relating to initial velocity(u), final velocity(v), time(t), acceleration(a) and displacement(S) are as follows:
(i) v = u + at
(ii) S = ut + ½ at^{2}
(iii) v^{2} = u^{2} +2As
2. Derive the following equations for a uniformly accelerated motion:
(i) v = u + at
(ii) S = ut + ½ at^{2}
(iii) v^{2} = u^{2} +2As
where the symbols have their usual meanings.
Solution
Derivation of equations of motion
First equation of motion:
Consider a particle moving along a straight line with uniform acceleration ‘a’. At t = 0, let the particle be at A and u be its initial velocity, and at t = t, let v be its final velocity.
Acceleration = Change in velocity/Time
a = (v-u)t
⇒ at = v-u
⇒ v = u + at ---(1)
Second equation of motion:
Distance travelled = average velocity × time
From equation (1), v=u + at, substituting the value of v in the above equation;
∴ S = ut + ½ at^{2} ---(2)
Third equation of motion:
Distance travelled = average velocity × time
v^{2} - u^{2} = 2as
= v^{2} = u^{2} + 2as
3. Write an expression for the distance S covered in time T by a body which is initially at rest and starts moving with a constant acceleration a.
Solution
distance = s
acceleration = a
time taken = t
initial velocity u=0
Therefore, S = ut + ½ at^{2}
Since, u = 0,
∴ S = ½ at^{2}
Multiple choice type:
1. The correct equation of motion is:
(a) v = u + 2aS
(b) v = ut + a
(c) S = ut+ ½ at
(d) v = u + at
Solution
(d) v = u + at
Where u is the initial velocity, v is the final velocity, a is the acceleration and t is the time
2. A car starting from rest accelerates uniformly to acquire a speed 20km/h in 30 min. The distance travelled by car in this time interval will be:
(a) 600 km
(b) 5 km
(c) 6km
(d) 10km
Solution
(b) 5 km
v = u + at
20 = 0 + a(1/2)
a= 40km/h
S = ut + ½ at^{2}
= 0(1/2) + ½ (40)(1/2)^{2}
= 0+5 = 5 km
Numericals
1. A body starts from rest with a uniform acceleration of 2m/s. Find the distance covered by the body in 2s.
Solution
Initial velocity u = 0
Acceleration a = 2 m/s^{2}
Time, t=2s
Let ‘S’ be the distance covered.
Using the second equation of motion,
S = ut + (1/2) at^{2}
S = 0 + (1/2) (2) (2)^{2}
S=4m
2. A body starts with an initial velocity of 10m/s and acceleration 5m/s’. Find the distance covered by it in 5s.
Solution
Initial velocity u = 10 m/s
Acceleration a = 5 m/s^{2}
Time t = 5s
Let ‘S‘ be the distance covered.
Using the second equation of motion,
S = ut + (1/2) at^{2}
S = (10)(5) + (1/2) (5) (5)^{2}
S = 50 + 62.5
S = 112.5m
3. A vehicle is accelerating on a straight road, Its velocity at any instant is 30km/h, after 2s, it is 33.6km/h and after further 2s, it is 37.2km/h. Find the acceleration of vehicle in m/s^{2}. Is the acceleration uniform?
Solution
Acceleration = Change in velocity/time taken
In the first two seconds,
Acceleration = (33.6 - 30)/2 km/h^{2}
= 1.8 km/h^{2}
= 0.5 m/s^{2} ...(i)
In the next two seconds,
Acceleration = (37.2 - 33.6)/2 km/h^{2}
= 1.8 km/h^{2}
= 0.5 m/s^{2} ...(ii)
From (i) and (ii), we can say that the acceleration is uniform.
4. A body, initially at rest, starts moving with a constant acceleration 2m/s^{2}. Calculate: (i) the velocity acquired and (ii) and the distance travelled in 5s.
Solution
Initial velocity u = 0 m/s
Acceleration a = 2 m/s^{2}
Time, t=5s
(i) Let ‘v’ be the final velocity.
Then, (v-u)/5 = 2
v=10m/s
(ii) Let ‘s’ be the distance travelled.
Using the third equation of motion,
v^{2}- u^{2} = 2as
We get,
10^{2} - 0^{2} = 2 ×2 ×s
Thus, s = (100/4) m = 25m
5. A body initially moving with a velocity 20m/s strikes a target and comes to rest after penetrating a distance 10 cm in the target. Calculate the retardation caused by the target.
Solution
Initial velocity u = 20 m/s
Final velocity v = 0
Distance travelled s = 10 cm = 0.1m
Let acceleration be ‘a’.
Using the third equation of motion,
v^{2} - u^{2} = 2as
We get,
0^{2} - 20^{2} = 2 ×a ×0.1
a = -(400/0.2) m/s^{2}
a = -2000 m/s^{2}
Thus, retardation = 2000 m/s^{2}
6. A train moving with a velocity of 20m/s is brought to rest by applying brakes in 5s. Calculate the retardation.
Solution
Initial velocity u = 20 m/s
Final velocity v = 0
Time taken, t = 5s
Let acceleration be ‘a’.
Using the first equation of motion,
v = u + at
0 = 20 + 5a
a = -4 m/s^{2}
Thus, retardation = 4 m/s^{2}
7. A train travels with a speed of 60km/h from station A to station B and then comes back with a speed 80km/h from station B to station A. Find: (i) the average speed, and (ii) the average velocity of train.
Solution
Let ‘s’ be the distance between stations A and B.
(i) Average speed = Total distance/total time taken
Here, total distance = s +s = 2s
Total time taken = Time taken to travel from A to B + Time taken to travel from B to A.
= [(s/60) + (s/80)]s
= [140s/4800]s
(ii) Average velocity = Displacement/total time taken
Because the train starts and ends at the same station, the displacement is zero. Thus the average velocity is zero.
8. A train is moving with a velocity of 90km/h. It is brought to stop by applying the brakes which produce a retardation of 0.5m/s^{2}. Find: (i) the velocity after 10s, and (ii) the time taken by the train to come to rest.
Solution
Initial velocity u = 90 km/h = 25 m/s
Final velocity v = 0 m/s
Acceleration a = -0.5 m/s^{2}
(i) Let ‘V’ be the velocity after time t = 10s
Using the first equation of motion,
v = u + at
We get,
v = 25 + (-0.5) (10) m/s
v = 25-5 = 20m/s
(ii) Let t’ be the time taken by the train to come to rest.
Using the first equation of motion,
v = u + at
We get,
t’ = [(0 - 25)/(-0.5)] s
t’=50s
9. A car travels a distance 100m with a constant acceleration and average velocity of 20m/s. The final velocity acquired by the car is 25m/s. Find: (i) the initial velocity and (ii) acceleration of car.
Solution
Distance travelled s = 100 m
Average velocity V = 20 m/s
Final velocity v = 25 m/s
(i) Let u be the initial velocity.
Average velocity = (Initial velocity + Final velocity)/2
v = (u+v)/2
⇒ 20 = (u + 25)/2
⇒ u = 40-25 = 15m/s
(ii) Let ‘a’ be the acceleration of the car.
Using the third equation of motion,
v^{2} - u^{2} = 2as
We get,
25^{2 }- 15^{2} = 2×a ×100
⇒ 625 - 225 = 200a
= (400/200) m/s^{2} = 2 m/s^{2}
10. When brakes are applied to a bus, the retardation produced is 25cm/s and the bus takes 20s to stop. Calculate: (i) the initial velocity of bus, and (ii) the distance travelled by the bus during this time.
Solution
Final velocity v = 0
Acceleration = -25 cm/s^{2} or -0.25 m/s^{2}
Time taken t = 20s
(i) Let ‘u’ be the initial velocity.
Using the first equation of motion,
v = u + at
We get,
u = v - at
⇒ u = 0 - (-0.25)(20) = 5 m/s
(ii) Let ‘s’ be the distance travelled.
Using the third equation of motion,
v^{2} - u^{2} = 2as
We get,
0^{2} - 5^{2} = 2 (-0.25) ×s
s = (25/0.5) = 50 m
11. A body moves from rest with a uniform acceleration and travels 270m in 3s. Find the velocity of the body at 10s after the start.
Solution
Initial velocity u = 0 m/s
Distance travelled s = 270m
Time taken to travel s distance = 3s
Let ‘a’ be the uniform acceleration.
Using the second equation of motion,
S = ut + ½ at^{2}
We get,
270 = 0 + ½ ×a ×3^{2}
⇒ 270 = 9a/2
⇒ a = 60 m/s^{2}
Let v be the velocity of the body 10 s after the start.
Using the first equation of motion, we get
v = u + at
v = 0 + 60 ×10 = 600 m/s
12. A body moving with a constant acceleration travels the distances 3m and 8m respectively in 1s and 2s. Calculate: (i) the initial velocity, and (ii) the acceleration of body.
Solution
Let the constant acceleration with which the body moves be ‘a’.
Given, the body travels distance S_{1} = 3 min time t_{1} = 1s.
Same body travels distance S_{2} = 8 min time t_{2} = 2 s.
(i) Let ‘u’ be the initial velocity.
Using the second equation of motion,
S = ut + ½ at^{2}
Substituting the value for S_{1} and S_{2}, we get
(ii) Putting u = 2 m/s in the equation
13. A car travels with a uniform velocity of 25m/s for 5s. The brakes are then applied and the car is uniformly retarded and comes to rest in further 10s, Find: (i) the distance which the car travels before the brakes are applied, (ii) the retardation and (iii) the distance travelled by the car after applying the brakes.
Solution
Initial velocity u = 25 m/s
Final velocity v = 0
(i) Before the brakes are applied, let S be the distance travelled.
Distance = Speed × time
S = 25×5 m
⇒ S = 125m
(ii) Acceleration = (Final velocity - Initial velocity)/Time taken
= [(0 - 25)/15] ms^{-2}
= (-5/2) ms^{-2}
= -2.5 ms^{-2}
Therefore, retardation = 2.5 ms^{-2}
(iii) After applying brakes, the time taken to come to stop = 10s
Let S’ be the distance travelled after applying the brakes.
Initial velocity u = 25 m/s
Final velocity v = 0
Using the third equation of motion,
v^{2} – u^{2} = 2as
We get,
0^{2} - 25^{2} = 2 ×-2.5 × S’
⇒ 625 = 5 × S’
⇒ S’= 125m
14. A space craft flying in a straight course with a velocity of 75km/s fires its rocket motors for 6.0s. At the end of this time, its speed is 120km/s in the same direction. Find: (i) the space craft’s average acceleration while the motors were firing, (ii) the distance travelled by the space craft in the first 10s after the rocket motors were started, the motors having been in action for only 6.0s.
Solution
Given, the initial velocity u = 75 km/s
Final velocity v = 120 km/s
Time taken = 6s
(i) Acceleration = (Final velocity - Initial velocity)/time taken
= [(120 - 75)/6] kms^{-2}
= (45/6) kms^{-2}
= 7.5 kms^{-2}
(ii) Distance travelled by the aircraft in the first 10 s = Distance travelled in the first 6 s + Distance travelled in the next 4 s.
Distance travelled in the first 6s (S_{1}) = ut + ½ at^{2}
(S_{1}) = ut + ½ at^{2}
(S_{1}) = (7 ×6) + (1/2) ×7.5 × 6^{2}
(S_{1}) = 450 + 135
(S_{1}) = 585 km
Distance travelled in the next 4s (S_{2}) = Speed × time
Speed at the end of 6 s is 120 km/s.
(S_{2}) = 120 ×4
(S_{2}) = 480 km
Thus, the distance travelled by the aircraft in the first 10 s = (S_{1}) + (S_{2}) = 585 + 480 = 1065 km.
15. A train starts from rest and accelerates uniformly at a rate of 2m/s^{2} for 10s. It then maintains a constant speed for 200s. The brakes are then applied and the train is uniformly retarded and comes to rest in 50s. Find: (i) the maximum velocity reached, (ii) the retardation in the last 50s, (iii) the total distance travelled, and (iv) the average velocity of the train.
Solution
(i) For the first 10 s, initial velocity u = 0
Acceleration a = 2 m/s^{2}
Time taken t = 10s
Let ‘v’ be the maximum velocity reached.
Using the first equation of motion
v = u+ at
We get
v = (0) + (2) (10) = 20 m/s
(ii) For the last 50 s: Final velocity = 0 m/s, initial velocity = 20 m/s.
Acceleration = (Final velocity - Initial velocity)/time
= (0 - 20)/50 = -0.4 m/s^{2}
Retardation = 0.4 m/s
(iii) Total distance travelled = Distance travelled in the first 10 s + Distance travelled in 200 s + Distance travelled in last 50s
Distance travelled in first 10s (S_{1}) = ut + ½ at^{2}
S_{1}= (0) + ½ (2) (10)^{2}
S_{1} = 100 m
Distance travelled in 200s (S_{2}) = speed x time
S_{2} = (20) (200) = 4000 m
Distance travelled in last 50s (S_{3}) = ut + ½ at^{2}
Here, u = 20 m/s, t = 50s and a = -0.4 m/s^{2}
S_{3}= (20)(50) + ½ (-0.4) (50)
S_{3} = 1000 - 500
S_{3} = 500 m
Therefore, total distance travelled = S_{1} + S_{2} + S_{3} = 100 + 4000 + 500 = 4600 m
(iv) Average velocity = Total distance travelled/total time taken
= (4600/260) m/s
= 17.69 m/s