CBSE Class 9 Physics Motion Worksheet

Fill in the blanks :

Question. when body covers equal distances in equal intervals of time then it is said to be having _____________________

Answer : Uniform

Question. Displacement has both direction and magnitude for its complete description and these physical quantities are called a _____

Answer : Vectors

Question. Distances between two cities are measured in ______________.

Answer : kilometres

Question. Basic unit of speed is __________.

Answer : metre per second (m/s)

Question. The to and fro motion of a simple pendulum is an example of a periodic or an _________

Answer : oscillatory motion.

Question. When a body moves from one position to another the shortest distance between the initial and final position of the body along with its direction is known as _______

Answer : Displacement

Question. The smallest time interval that can be measured with commonly available clocks and watches is____________.

Answer : one second

Question. Speedometer is an instrument that indicates the ____________.

Answer : speed of a vehicle.

Question. when the object covers unequal distances in equal intervals or equal distances in unequal intervals then motion is said be having ____ motion.

Answer : Non Uniform

Question. __________________ and ______ are used to describe the overall motion of an object and to locate its final position with reference to its initial position at a given time.

Answer : Distance, Displacement

Question. An object moving along a straight line with a constant speed is said to be in ____________.

Answer : uniform motion

Question. The distance travelled by a moving object cannot be zero but the final _______ of a moving object can be zero.

Answer : Displacement

Question. Physical quantities that do not require direction for their complete description are called ______

Answer : Scalars

Question. The distance moved by an object in a unit time is called its_____________.

Answer : speed

Question. If the speed of an object moving along a straight line keeps changing, its motion is said to be ________________.

Answer : non-uniform.

Short Answer Type Questions :

Question. Distinguish between speed and velocity.
Answer :

Question. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?
Answer : Uniform Acceleration : In this type of acceleration, the body moves along a straight line and its velocity increases/decreases at a uniform rate (it changes at a constant rate in any constant time interval).
Non-Uniform Acceleration : In this type of acceleration, the body moves along a straight line and its velocity increases/decreases at a rate that is not uniform (it changes at a different rate for a given constant time interval).

Question. What does the path of an object look like when it is in uniform motion?
Answer : The path of an object in uniform motion is a straight line.

Question. Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object.
Answer : Neither of the statements are true. Statement (a) is false because the displacement of an object which travels a certain distance and comes back to its initial position is zero. Statement (b) is false because the displacement of an object can be equal to, but never greater than the distance traveled.

Question. A trolley, while going down an inclined plane, has an acceleration of 2 cm s^-2. What will be its velocity 3 s after the start?
Answer : Given, initial velocity (u) = 0 (the trolley begins from the rest position)
Acceleration (a) = 0.02 ms^-2
Time (t) = 3s
As per the first motion equation,
Therefore, terminal velocity of the trolley (v) = 0 + = 0.06 ms^-2
Therefore, the velocity of the trolley after 3 seconds will be 6 cm.s^-2

Question. What is the quantity which is measured by the area occupied below the velocity-time graph?
Answer : Considering an object in uniform motion, its velocity-time graph can be represented as follows.

Now, the area below the velocity-time graph is the area of the rectangle OABC, which is given by OA*OC. But OA is the velocity of the object and OC represents time. Therefore, the shaded area can be represented as:
Area under the velocity-time graph = velocity*time. Substituting the value of velocity as in the previous equation, it is found that the area under the velocity-time graph represents the total displacement of the object.

Question. A racing car has a uniform acceleration of 4 m s^-2. What distance will it cover in 10 s after start?
Answer : Given, the car is initially at rest; initial velocity (u) = 0 ms^-1
Acceleration (a) = 4 ms^-2
Time period (t) = 10 s
As per the second motion equation,
Therefore, the total distance covered by the car (s) =
= 200 meters
Therefore, the car will cover a distance of 200 meters after 10 seconds.

Question. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Answer : For uniform motion, the distance-time graph is a straight line. On the other hand, the distance-time graph of an object in non-uniform motion is a curve.

The first graph describes uniform motion and the second one describes non-uniform motion.

Question. What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
Answer : This speed-time graph can be plotted as follows.

Since there is no change in the velocity of the object (Y-Axis value) at any point of time (X-axis value), the object is said to be in uniform motion.

Question. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h–1 in 10 minutes. Find its acceleration.
Answer : Given, the initial velocity (u) of the train = 0m.s^-1 (at rest)
Terminal velocity (v) of the train = 40km/hour = 11.11 m.s^-1
Time interval, t = 10 minutes = 600 s.
The acceleration of the train is given by = = 0.0185 m.s^-2

Question. A train is travelling at a speed of 90 km h^–1. Brakes are applied so as to produce a uniform acceleration of –0.5 m s^-2. Find how far the train will go before it is brought to rest.
Answer : Given, initial velocity (u) = 90 km/hour = 25 m.s^-1
Terminal velocity (v) = 0 m.s^-1
Acceleration (a) = -0.5 m.s^-2
As per the third motion equation, Therefore,
distance travelled by the train (s) = s
= meters = 625
meters The train must travel 625 meters at an acceleration of -0.5 ms^-2 before it reaches the rest position.

Question. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Answer : This distance-time graph can be plotted as follows.

Since there is no change in the distance traveled by the object (or the Y-Axis value) at any point in the X-Axis (time), the object is at rest.

Question. A bus decreases its speed from 80 km h–1 to 60 km h–1 in 5 s. Find the acceleration of the bus.
Answer : Given, the initial velocity (u) = 80km/hour = = 22.22 m.s-1
The final velocity (v) = 60km/hour = = 16.66 m.s^-1
Time frame, t = 5 seconds.
Therefore, acceleration (a) = =
= -1.112 m.s-2
Therefore, the total acceleration of the bus is -1.112m.s^-2. It can be noted that the negative sign indicates that the velocity of the bus is decreasing. 

Question. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer : Yes, an object moving a certain distance can have zero total displacement. Displacement refers to the shortest distance between the initial and the final positions of the object. Even if an object moves through a considerable distance, if it eventually comes back to its initial position, the corresponding displacement of the object would be zero.

Question. What does the odometer of an automobile measure?
Answer : The odometer measures the total distance traveled by the automobile.

Question. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Answer : Since average speed is the total distance travelled in a time frame and velocity is the total displacement in the time frame, the magnitude of average velocity and average speed will be the same when the total distance travelled is equal to the displacement.

Question. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 10⁸ m/s.
Answer : Given that the signal travels in a straight line, the distance between the spaceship and the ground station is equal to the total distance traveled by the signal.
5 minutes = 5*60 seconds = 300 seconds. Speed of the signal = 3 × 10⁸ m/s.
Therefore, total distance = (3 × 10⁸ m/s) * 300s
= 9*10¹⁰ meters.

Long Answer Type Questions :

Question. A farmer moves along the boundary of a square field of side 10m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
Answer : Given that the farmer covers the entire boundary of the square field in 40 seconds, the total distance traveled by the farmer in 40 seconds is 4*(10) = 40 meters.
Therefore, the average distance covered by the farmer in one second is:
Two minutes and 20 seconds can be written as 140 seconds. The total distance traveled by the farmer in this timeframe is:
Since the farmer is moving along the boundary of the square field, the total number of laps completed by the farmer will be:
Now, the total displacement of the farmer depends on the initial position. If the initial position of the farmer is at one corner of the field, the terminal position would be at the opposite corner (since the field is square).
In this case, the total displacement of the farmer will be equal to the length of the diagonal line across the opposite corners of the square.
Applying the Pythagoras theorem, the length of the diagonal can be obtained as follows: = = 14.14m.
This is the maximum possible displacement of the farmer.
If the initial position of the farmer is at the mid-point between two adjacent corners of the square, the net displacement of the farmer would be equal to the side of the square, which is 10m. This is the minimum displacement.
If the farmer starts at a random point around the perimeter of the square, his net displacement after traveling 140m will lie between 10m and 14.14m.

Question. A bus starting from rest moves with a uniform acceleration of 0.1 m s^-2 for 2 minutes.
Find (a) the speed acquired,
(b) the distance travelled.
Answer : (a) Given, the bust starts from rest. Therefore, initial velocity (u) = 0 m/s
Acceleration (a) = 0.1 m.s^-2
Time = 2 minutes = 120 s
Acceleration is given by the equation
Therefore, terminal velocity (v) =
= (0.1 m.s^-2 * 120s) + 0 m.s^-1
= 12m.s^-1 + 0 m.s^-1
Therefore, terminal velocity (v) = 12m/s
(b) As per the third motion equation,
Since a = 0.1 m.s^-2, v = 12 m.s^-1, u = 0 m.s^-1, and t = 120s, the following value for s (distance) can be obtained.
Distance, s = ut + ½(at²)
Therefore, s = 720m.
The speed acquired is 12m.s^-1 and the total distance travelled is 720m.

Question. A stone is thrown in a vertically upward direction with a velocity of 5 m s^-1. If the acceleration of the stone during its motion is 10 m s^-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Answer : Given, initial velocity (u) = 5 m/s
Terminal velocity (v) = 0 m/s (since the stone will reach a position of rest at the point of maximum height)
Acceleration = 10 ms^-2 in the direction opposite to the trajectory of the stone = -10 ms^-2
As per the third motion equation,
Therefore, the distance traveled by the stone (s) =
Distance (s) = 1.25 meters
As per the first motion equation,
Therefore, time taken by the stone to reach a position of rest (maximum height) =
= Time taken = 0.5 seconds
Therefore, the stone reaches a maximum height of 1.25 meters in a timeframe of 0.5 seconds.

Multiple Choice Questions

1. If the displacement of an object is proportional to square of time, then the object moves with
(a) uniform velocity
(b) uniform acceleration
(c) increasing acceleration
(d) decreasing acceleration

2. The distance time graph of a body coincides with its time axis. The body must be
(a) in uniform motion
(b) at rest
(c) in uniformly accelerated motion
(d) in zig-zag motion

3. From the given v – t graph (see below Fig.), it can be inferred that the object is
(a) in uniform motion
(b) at rest
(c) in non-uniform motion
(d) moving with uniform acceleration

4. The velocity time graph of a body is parallel to the time axis. The body is
(a) at rest
(b) having uniform acceleration
(c) having zero acceleration
(d) having non-uniform acceleration

5. A particle is moving in a circular path of radius r. The displacement after half a circle would be:
(a) Zero
(b) π r
(c) 2 r
(d) 2π r

6. A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is,
(a) u/g
(b) u²/2g
(c) u²/g
(d) u/2g

7. The numerical ratio of displacement to distance for a moving object is
(a) always less than 1
(b) always equal to 1
(c) always more than 1
(d) equal or less than 1

8. Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of 10 m/s. It implies that the boy is
(a) at rest
(b) moving with no acceleration
(c) in accelerated motion
(d) moving with uniform velocity

9. Area under a v – t graph represents a physical quantity which has the unit
(a) m²
(b) m
(c) m³
(d) m/s

10. Four cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown in below Fig.. Choose the correct statement

(a) Car A is faster than car D.
(b) Car B is the slowest.
(c) Car D is faster than car C.
(d) Car C is the slowest.

11. Slope of a velocity – time graph gives
(a) the distance
(b) the displacement
(c) the acceleration
(d) the speed

12. In which of the following cases of motions, the distance moved and the magnitude of displacement are equal?
(a) If the car is moving on straight road
(b) If the car is moving in circular path
(c) The pendulum is moving to and fro
(d) The earth is revolving around the Sun

13. Which of the following figures (see below Figure) represents uniform motion of a moving object correctly?

SHORT ANSWER QUESTIONS

14. The displacement of a moving object in a given interval of time is zero. Would the distance travelled by the object also be zero? Justify you answer.

15. How will the equations of motion for an object moving with a uniform velocity change?

16. A car starts from rest and moves along the x-axis with constant acceleration 5 m/s2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest?

17. A motorcyclist drives from A to B with a uniform speed of 30 km/h and returns back with a speed of 20 km/h. Find its average speed.

18. Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height.

19. The velocity-time graph (see below Figure) shows the motion of a cyclist. Find (i) its acceleration (ii) its velocity and (iii) the distance covered by the cyclist in 15 seconds.

20. A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement–time graph is shown in below figure. Plot a velocity– time graph for the same.

LONG ANSWER QUESTIONS

21. An object starting from rest travels 20 m in first 2 s and 160 m in next 4 s. What will be the velocity after 7 s from the start.

22. An electron moving with a velocity of 5 × 104 m/s enters into a uniform electric field and acquires a uniform acceleration of 104 m/s2 in the direction of its initial motion.
(i) Calculate the time in which the electron would acquire a velocity double of its initial velocity.
(ii) How much distance the electron would cover in this time?

23. Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds.

24. Two stones are thrown vertically upwards simultaneously with their initial velocities u¹ and u² respectively. Prove that the heights reached by them would be in the ratio of u²₁ :u₂² (Assume upward acceleration is –g and downward acceleration to be +g).

25. An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2 s if both the objects drop with same accelerations? How does the difference in heights vary with time?

1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Solution.
Given, diameter of the track (d) = 200m
Therefore, circumference of the track (π*d) = 200π meters.
Distance covered in 40 seconds = 200π meters
Distance covered in 1 second =
Distance covered in 2minutes and 20 seconds (140 seconds) = meters
= meters = 2200 meters
Number of laps completed by the athlete in 140 seconds = = 3.5
Therefore, the final position of the athlete (with respect to the initial position) is at the opposite end of the circular track. Therefore, the net displacement will be equal to the diameter of the track, which is 200m.
Therefore, the net distance covered by the athlete is 2200 meters and the total displacement of the athlete is 200m.

2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Solution.
Given, distance covered from point A to point B = 300 meters
Distance covered from point A to point C = 300m + 100m = 400 meters
Time taken to travel from point A to point B = 2 minutes and 30 seconds = 150 seconds
Time taken to travel from point A to point C = 2 min 30 secs + 1 min = 210 seconds
Displacement from A to B = 300 meters
Displacement from A to C = 300m – 100m = 200 meters
Average speed =
Average velocity =
Therefore, the average speed while traveling from A to B = = 2 m/s
Average speed while traveling from A to C = = 1.9 m/s
Average velocity while traveling from A to B = = 2 m/s
Average velocity while traveling from A to C = = 0.95 m/s

3. Abdul, while driving to school, computes the average speed for his trip to be 20 km.h^–1. On his return trip along the same route, there is less traffic and the average speed is 30 km.h^–1. What is the average speed for Abdul’s trip?
Solution.
Distance travelled to reach the school = distance travelled to reach home = d (say)
Time taken to reach school = t₁
Time taken to reach home = t₂ therefore, average speed while going to school = 20 km/h
Average speed while going home = 30 km/h
Therefore, and
Now, the average speed for the entire trip is given by
= km/h = km/h
= kmh^-1
= 120/5 kmh^-1 = 24 kmh^-1
Therefore, Abduls average speed for the entire trip is 24 kilometres per hour.

4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s–2 for 8.0 s. How far does the boat travel during this time?
Solution.
Given, initial velocity of the boat = 0 m/s
Acceleration of the boat = 3 ms-2
Time period = 8s
As per the second motion equation,
Therefore, total distance travelled by the boat in 8 seconds =
= 96 meters
Therefore, the motorboat travels a distance of 96 meters in a time frame of 8 seconds.

5. A driver of a car travelling at 52 km h–1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h^–1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
Solution.
The speed v/s time graphs for the two cars can be plotted as follows.

The total displacement of each car can be obtained by calculating the area beneath the speed-time graph.
Therefore, displacement of the first car = area of triangle AOB
= (1/2)*(OB)*(OA)
But OB = 5 seconds and OA = 52 km.h^-1 = 14.44 m/s
Therefore, the area of the triangle AOB is given by: (1/2)*(5s)*(14.44ms^-1) = 36 meters
Now, the displacement of the second car is given by the area of the triangle COD
= (1/2)*(OD)*(OC)
But OC = 10 seconds and OC = 3km.h^-1 = 0.83 m/s
Therefore, area of triangle COD = (1/2)*(10s)*(0.83ms^-1) = 4.15 meters
Therefore, the first car is displaced by 36 meters whereas the second car is displaced by 4.15 meters. Therefore, the first car (which was traveling at 52 km/h) travelled farther post the application of brakes.

6. Fig 8.11 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 C travelled when B passes A? (d) How far has B travelled by the time it passes C?
Solution.
(a) since the slope of line B is the greatest, B is traveling at the fastest speed.
(b) since the three lines do not intersect at a single point, the three objects never meet at the same point on the road.
(c) since there are 7 unit areas of the graph between 0 and 4 on the Y axis, 1 graph unit equals 4/7 km.
Since the initial point of object C is 4 graph units away from the origin, Its initial distance from the origin is 4*(4/7)km = 16/7 km
When A passes B, the distance between the origin and C is 8km
Therefore, total distance travelled by C in this time = 8 – (16/7) km = 5.71 km
(d) the distance that object B has covered at the point where it passes C is equal to 9 graph units.
Therefore, total distance travelled by B when it crosses C = 9*(4/7) = 5.14 km

7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s^-2, with what velocity will it strike the ground? After what time will it strike the ground?
Solution.
Given, initial velocity of the ball (u) = 0 (since it began at the rest position)
Distance travelled by the ball
As per the third motion equation,
Therefore,
= 2*(10ms^-2)*(20m) + 0
v2 = 400m2s-2
Therefore, v= 20ms^-1
The ball hits the ground with a velocity of 20 meters per second.
As per the first motion equation,
Therefore,
=
= 2 seconds
Therefore, the ball reaches the ground after 2 seconds.

8. The speed-time graph for a car is shown is Fig. 8.12

(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?
Solution.
(a)

The shaded area represents the displacement of the car over a time period of 4 seconds. It can be calculated as:
(1/2)*4*6 = 12 meters. Therefore the car travels a total of 12 meters in the first four seconds.
(b) Since the speed of the car does not change from the points (x=6) and (x=10), the car is said to be in uniform motion from the 6th to the 10th second.

9. 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.
Solution.
(a) It is possible; an object thrown up into the air has a constant acceleration due to gravity acting on it. However, when it reaches its maximum height, its velocity is zero.
(b) it is impossible; acceleration implies an increase or decrease in speed, and uniform speed implies that the speed does not change over time
(c) It is possible; for an object accelerating in a circular trajectory, the acceleration is perpendicular to the direction followed by the object.

10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Solution.
Given, radius of the orbit = 42250 km
Therefore, circumference of the orbit = 2*π*42250km = 265571.42 km
Time taken for the orbit = 24 hours
Therefore, speed of the satellite = 11065.4 km.h^-1
The satellite orbits the Earth at a speed of 11065.4 kilometres per hour.

More Question

1. What is meant by the statement ‘Rest and motion are relative terms’? Give example to show it.

2. Explain whether the walls of a classroom are at rest or in motion.

3. Define scalar and vector quantities.

4. Identify the following as scalar or vector quantities:- mass, velocity, speed, length, distance, displacement, temperature, force, weight, power, work and energy.

5. The school of a boy from his home is 1 km to the east. When he reaches back home, he says that he had traveled 2 km distance but his displacement is zero. Justify your answer.

6. Under what condition, the average speed is equal to the magnitude of the average velocity.

7. Can the average speed of a moving body be zero?

8. Can the average velocity of a moving body be zero? State eamples.

9. A car covers a distance of 5 km in 20 mins. Find the velocity of the car in (a) km/min (b)m/s (c) m/min (d) km/hr.

10. a train is moving with a velocity of 45km/hr. calculate the distance traveled by it in 1 hr, 1 min, 1 second.

11. An object P is moving with a constant velocity for 5 mins. Another object Q is moving with changing velocity for 5 mins. Out of these two objects, which one has acceleration? Explain.

12. Can an object be accelerated if it is moving with constant speed? If yes, explain giving examples.

13. (i) When do you say that an object has positive acceleration?
(ii) When do you say that an object has negative acceleration?

14. State which of the following situations are possible and give an example of each of these:-
(a) a body moving with constant acceleration but with zero velocity.
(b) A body moving horizontally with acceleration in vertical direction.
(c) A body moving with a constant speed in an accelerated motion.

15. What is a reference point?

16. Name the 2 physical quantities which can be obtained from velocity-time graph.

17. An electric train is moving with a velocity of 120km/hr. how much distance will it cover in 30 sec?

18. Give differences between linear motion and circular motion.

19. Velocity time graph of a body is shown in the figure. What are initial and final velocities of the body?

20. A body moves around the sun with constant speed in circular path. Is the motion of the bodyuniform or accelerated?

21. Name the physical quantity which remains constant during uniform circular motion.

22. Name the physical quantity which changes during uniform circular motion.

23. An object has moved through a distance. Can it have zero displacement? Support your answer with an example.

24. A physical quantity is measured – 10m/s. is it speed or velocity?

25. A car is moving with a uniform velocity of 10m/s. the driver of the car decides to overtake the bus moving ahead of the car. So the driver of the car accelerates at 1m/s2 for 10 sec. Find the velocity of the car at the end of 10 sec. also find the distance traveled by the car while accelerating

Section A: Conceptual and Application Questions

1. What is the condition under which the distance and the magnitude of displacement are equal?

2. What is the numerical ratio of average velocity to average speed of an object when it is moving along a straight path?

3. Is it possible for distance- time graph of a moving particle to be perpendicular to time axis? Justify your answer.

4. Is it possible that the average velocity of a particle is zero when its average speed is not zero?

5. Give two points of difference between speed and velocity. How is average speed of an object calculated when its speed changes?

6. Draw the displacement- time graphs for an object (i) at rest (ii) in uniform motion (iii) in uniformly accelerated motion.

7. Derive graphically (i) position-time relation (ii) velocity – time relation for an object under uniformly accelerated motion.

8. Can uniform circular motion be considered as an accelerated motion? Explain.

9. The slope of velocity –time graph represents which physical quantity? Write its S I unit.

10. The displacement of a moving object in a given time interval is zero. Would the distance travelled by the object also be zero? Justify your answer.

Section B: Numerical Questions

11. The minute hand of a clock is 7 cm long. Find the displacement and distance covered by the minute hand from 9 am to 9.30 am.

12. A car covers 30 km at a uniform speed of 30 km/h. What would be the speed of the car for the next 90 km if the average speed for the entire journey is 60 km/h?

13. Rajeev went from Delhi to Chandigarh on his motorbike. The odometer of the bike read 4200 km at the start of trip and 4460 km at the end of his trip. If Rajeev took 4 hours 20 minutes to complete his trip, find the average speed in kmh-1 as well as in ms^-1.

14. A cyclist goes around a circular track once every 2 minutes. If the radius of the circular track is 110 m, calculate the speed of the cyclist.

15. Which type of motion is represented by the velocity- time graph given below?

(a) Name the physical quantity which can be calculated by the area of rectangle OABC.
(b) What does the straight line AB represent?

16. A scooter moving at a speed of 10 m/s is stopped by applying brakes which produce a uniform acceleration 0.5 m/s2 .How much distance will be covered by the scooter before it stops?

17. A train travelling at a speed of 36 km/h is brought to rest by applying brakes. If it travels 100 m before coming to rest, find the value of retardation.

18. An object starting from rest travels 20 m in first 2 s and 160 m in next 4 s. Assuming uniform acceleration, what will be the velocity after 7 s from the start?

19. A body starting from rest travels with uniform acceleration. If it travels a distance of 100 m in 5 s. Find the value of acceleration.

20. Study the velocity – time graph of an ascending passenger lift in the figure shown below.

(a) What is the acceleration of the lift during:
(i) the first two seconds
(ii) between second and tenth second
(iii) during the last two seconds
(b) Which physical quantity is measured by area under the quadrilateral OABC? Calculate that physical quantity.

21. A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming the acceleration is uniform, find (i) the acceleration (ii) the distance travelled by the train for attaining this velocity.

22. A motorcyclist riding motorcycle A, who is traveling at a speed of 20 m/s applies the brakes and stops the motorcycle in 10s. Another motorcyclist of motorcycle B who is traveling at 5 m/s applies the brakes and stops the motorcycle in 20s. Plot speed-time graph for the two motorcycles. Which of the two motorcycles traveled farther before it comes to a stop?

23. A car starts from rest and moves along the x-axis with constant acceleration 5 m s–2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest?

24. The body moves with a velocity of 2 m/s for 5 s, and then its velocity increases uniformly to 10 m/s in next 5s. Thereafter its velocity begins to decreases at a uniform rate until it comes to rest after 5s.
(a) plot the velocity time graph for the motion of the body
(b) from the graph find the total distance covered by the body after 2 s and 12 s.

25. Two stones are thrown vertically upwards simultaneously with their initial velocities u₁ and u₂ respectively. Prove that the heights reached by them would be in the ratio of u₁² : u²₂. Assume upward acceleration is –g and downward acceleration is +g.

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1. A train travels 40 km at a uniform speed of 30 km h-1. Its average speed after travelling another 40 km is 45 km h^-1 for the whole journey. Its speed in the second half of the journey is
(A) 45 km h^-1
(B) 90 km h^-1
(C) 60 km h^-1
(D) None of these

2. 

In the above velocity-time graph of a moving object
(A) Acceleration in the first 2 seconds is 2.3 ms-1
(B) Acceleration in the last 2 seconds is -2.3ms-1
(C) Motion is uniform between second and tenth second
(D) All are correct

3. In the given velocity-time graph AB shows that the body has

(A) Uniform acceleration
(B) Uniform deceleration
(C) Uniform velocity throughout the motion
(D) None of these

4. In the adjoining velocity-time graph

(A) Velocity is decreasing with time
(B) Deceleration is uniform
(C) The object has a negative velocity beyond the point B
(D) All are correct

5. A train starting from rest attains the velocity of 72 km h-1. If the train has taken 5 minutes, it would have travelled a distance of
(A) 15 km
(B) 3 km
(C) 6 km
(D) None of these

6. An object has travelled 10 km in 15 minutes, its displacement will be
(A) 10 km
(B) Can be zero
(C) More than 10 km
(D) Cannot be predicted

7. Define Centripetal force.

8. When do we say the acceleration of a body is zero?

9. What is meant by non-uniform acceleration?

10. What is meant by uniform acceleration?

11. Define acceleration?

12. When does the velocity and speed of a moving body become identical?

13. What is meant by average velocity of a body moving in a particular direction?

14. When does the velocity change?

15. What is meant by the displacement?

16. What is meant by the term distance?

17. What is a scalar quantity?

18. How is the position of an object described?

19. What is meant by body in motion?

20. What is meant by body at rest?

21. Why is uniform linear motion not an accelerated motion?

22. A moving train is brought to rest within 20 sec by applying brakes. Find the initial velocity, if the retardation due to brakes is 2 m/s2 .

23. What is positive acceleration and negative acceleration?

24. What is meant by non-uniform velocity?

25. What is meant by uniform velocity (constant velocity) ?

26. Define velocity?

27. What is non-uniform motion?

28. What is uniform motion?

29. What is a vector quantity?

30. A jeep starts from rest and attains a speed of 40 km h-1 in 10 minutes. The uniform acceleration will be
(A) 4 km h^-1
(B) 4 km m^-1
(C) 66.7 m s^-2
(D) 66.7 m s^-1

31. Displacement is the
(A) Shortest distance between initial and final positions
(B) The actual distance between initial and final positions
(C) The distance travelled by the object
(D) Distance travelled by the object in a unit time

32. This time-displacement graph shows

(A) An object moving with a uniform velocity
(B) An object moving with a non-uniform speed
(C) A uniformly accelerated motion
(D) No information

33. Motion of an object is the change in position with respect to a reference point, also known as
(A) Origin
(B) Initial position
(C) Final position
(D) Distance

34. If an object covers equal distances in equal intervals of time, it is said to be in
(A) Circular Motion
(B) Uniform Motion
(C) Oscillatory Motion
(D) Non-uniform Motion

35. Average speed of an object is obtained by
(A) Dividing the total distance travelled by the total time taken
(B) Half of the sum of the initial speed and the final speed
(C) Both (1) and (2)
(D) Multiplying total distance with the total time.

36. S.I. Unit of speed is metre per second. Other units of speed are
(A) km h^-1
(B) cm s^-1
(C) h Km^-1
(D) Only (1) and (2)

37. Speed is a
(A) Scalar quantity
(B) Vector quantity
(C) Fixed quantity
(D) Unequal quantity

38. Velocity is the speed of an object moving
(A) In a definite direction
(B) In any direction
(C) In circular direction
(D) Reverse direction

39. Average velocity of an object is obtained by
(A) Dividing the total distance travelled by the total time taken
(B) Half of the sum of the initial velocity and the final velocity
(C) Both (1) and (2)
(D) None of the above

40. Magnitude of average speed of an object is equal to its average velocity if
(A) It is moving in a definite direction
(B) Its initial and final positions are same
(C) It is a uniform motion
(D) None of these

41. Acceleration of an object is
(A) The change in velocity per unit time
(B) A vector quantity
(C) In the direction of velocity
(D) All of the above

42. Negative value of acceleration signifies
(A) The velocity is increasing
(B) The velocity is decreasing
(C) The velocity remains the same
(D) The object comes to rest

43. In distance-time graphs
(A) Distance is taken along the X- axis
(B) Time is taken along the Y-axis
(C) Straight line indicates uniform motion
(D) Straight line indicates non-uniform motion

44. In velocity-time graphs
(A) Velocity is taken along the Y-axis and time is taken along the X-axis
(B) Straight line indicates uniform acceleration
(C) Straight line parallel to x-axis indicates uniform motion
(D) All of the above

45. In the velocity-time graphs, the distance is given by
(A) The area enclosed by the velocity-time graph and x-axis
(B) The length of the graph line
(C) The slope of the line
(D) None of these

46. The equations of motion can be represented as
(A) v = u + at
(B) s = ut + ½ at²
(C) 2as = v² - u²
(D) All of these

47. Motion of a satellite in circular orbit is an example of
(A) Uniform circular motion
(B) Accelerated circular motion
(C) Non-uniform circular motion
(D) Linear motion

48. Average speed of a car is 50 km h^-1. It can also be expressed in S.I. Units as
(A) 13.9 m s^-1
(B) 5 m s^-1
(C) 50 m s^-1
(D) 139 m s^-1

49. A man moved 6 m in south and then turned towards east to move 8 m. What is the distance travelled by the man and the displacement of the man?
(A) Distance= 6 m, Displacement= 8 m
(B) Distance= 14 m, Displacement= 8 m
(C) Distance= 14 m, Displacement= 10 m
(D) Distance= 14 m, Displacement= 14 m

50. Study the following table:
Distance         Time taken
10 Km                2 hr
25 Km                4 hr
60 Km                6 hr
What type of motion is exhibited by the body?
(A) Uniform motion
(B) Non-uniform motion
(C) Uniformly accelerated motion
(D) Non-uniformly accelerated motion

51. Convert Km/h² into m/s²:
(A) 5/18
(B) 5/36
(C) 5/54
(D) 5/324

52. Which physical quantity varies in a uniform circular motion?
(A) Speed
(B) Velocity
(C) Acceleration
(D) Both velocity and acceleration

53. Two cars A and B covered 40 Km and 60 Km in 2 hours and 3 hours respectively. Which of the following is moving faster?
(A) Car A
(B) Car B
(C) Both are moving with same velocity
(D) Cannot be predicted from the given data

54. The value of acceleration if velocity-time graph is a straight line parallel to time axis is:
(A) Infinity
(B) Zero
(C) One
(D) Cannot be predicted

55. Which of the following situation/situations is/are not possible?
(i) Velocity-time graph with a line parallel to velocity axis.
(ii) Velocity-time graph with downward slope.
(iii) Distance-time graph with a line parallel to time axis.
(A) Statement (i) only
(B) Statement (ii) only
(C) Both statements (i) and (ii)
(D) Statements (ii) and (iii)

56. Area under velocity time graph indicates:
(A) Magnitude of displacement
(B) Magnitude of acceleration
(C) Both magnitude and direction of displacement
(D) Both magnitude and direction of acceleration

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57. What is meant by angular acceleration?

58. What is meant by angular velocity? Write its unit.

59. Why is the motion of Satellites around their planets considered an accelerated motion?

60. A racing car has a uniform acceleration of 4 ms^-2. What distance will it cover in 10 seconds after the start?

61. A bus covers a distance of 250 km from Delhi to Jaipur towards West in 5 hours in the morning and returns to Delhi in the evening covering the same distance of 250km in the same time of 5 hours. Find (a) average speed & (b) average velocity of the bus for the whole journey.

62. A car travels a distance of 200 km from Delhi to Ambala towards North in 5 hours.
Calculate (i) Speed & (ii) Velocity of car for this journey?

63. What are equations of motion?

64. A body thrown vertically upward rises up to a height H, and comes back to the initial position.
i) Calculate the total distance travelled by the body
ii) The displacement of the body

65. The train A travelled a distance of 120 km in 3 hours where as another train B travelled a distance of 180 Km in 4 hours. Which train travelled faster.

66. Explain the uniform and non uniform motion by distance time graph.

67. Convert a speed of 72km/hr into
a) m/s b) cm/s

68. What are the characteristics of scalar quantities?

69. A train starting from rest attains a velocity of 72km h^-1 in 5 min. Assuming that the acceleration is uniform; find
(i) The acceleration.
(ii) The distance travelled by the train for attaining this velocity.

70. A cyclist goes around a circular track once every 2 minutes. If the radius of the circular track is 105 metres, calculate his speed. (Given π= 22/7)

71. Why is uniform circular motion called accelerated motion?

72. A scooter moving at a speed of 10m/s is stopped by applying brakes which produce a uniform acceleration of -0.5 ms-2 .How much distance will be covered by the scooter before it stops?

73. A driver change the speed of car from 25m/s to 10m/s in 5 seconds. Find the acceleration of the car.

74. The average speed of a bicycle, an athlete and a car are 18km/hr, 7m/s and 2 km/min respectively. Which of these is the fastest and which of these is slowest?

75. What are the characteristics of displacement?

76. What are the characteristics of vector quantities?

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77. The brakes applied to a car produced an acceleration of 6 ms^-2 in the opposite direction of the motion. If the car takes 2s to stop after the application of brakes, calculate the distance it travels during this time.

78. A scooter acquires a velocity of 36km per hour in 10 seconds just after the start.
Calculate the acceleration of the scooter.

79. Consider the situation shown in fig. below:
a) What is the position of a particle when it is at P₁ and when it is at P₂.
b) Are the positions same?
c) Are the two distance of the particle from the origin same?

80. A car covers 30km at a uniform speed of 60km/h and the next 30 km/h at a uniform speed of 40 km/h. find the total time taken.

81. A bus between Vishakhapatnam and Hyderabad passed the 100km,
160km and 220km points at 10.30 a.m., 11.30 a.m. and 1.30. p.m.
Find the average speed of the bus during each of the following intervals.

82. A train travels at a speed of 60km/hr for 0.52/hr, at 30km/h for the next 0.24 hr and then at 70km/h for the next 0.71h.What is the average speed of the train?

83. A man travels a distance of 1.5m towards East, then 2.0m towards South and finally 4.5m towards East.
i) What is the total distance traveled?
ii) What is his total displacement?

84. Define the following
a). Speed.
b). Average speed.
c). Uniform speed.

85. A particle is moving in a circle of radius . Show the following position of the particle:
a). from the center,300North-East.
b). from the center, 300West-North.
c). from the center towards South.

86. Differentiate between:-
a) Speed and average speed
b) Speed and velocity
c) Uniform linear motion and uniform circular motion

87. How many different types of velocity-time (speed-time) graphs are possible for a straight-line motion.

88. Explain using distance - time graphs:
a) When the body is at rest.
b) When the body is moving with a uniform speed
c) When the body is moving with a non-uniform speed.

89. Write the mathematical expression & S.I. units of the following?
1. Speed
2. Average velocity
3.Velocity
4. Average Velocity
5. Acceleration

90. A body falls freely. What is constant .

91. A body starts from rest. What is zero ?

92. A body stops after some time. What is zero ?

93. Displacement of a body is 3 metre when distance travelled by it is 2 metre. Can it be true ?

94. Can a body have acceleration when its velocity is zero ? ..

95. A body moving with uniform speed may have variable velocity. Exemplify.

96. What is the nature of displacement time graph of a body moving uniformly along a straight line ?

97. How do you calculate velocity of a body from its displacement time graph ?

98. The displacement time graph of a body is parallel to time axis. What does it signify ?

99. The displacement time graph of a body is parallel to displacement axis. Is it true ?

100. What is represented by slope of velocity time graph ?

101. Velocity time graph of a body is parallel to time axis. What is acceleration of the body ?

102. What is nature of velocity time graph of a body moving with uniform retardation ?

103. Displacement can be zero when distance travelled is not zero. But when distance travelled is zero, displacement must be zero. Justify.

104. A particle is travelling in a circle of diameter 15 metre. Calculate the distance covered and the displacement when it completes two rounds.

105. The speed of a car increases from 18 km/h to 36 km/h in 10 second. What is its acceleration?

106. Draw velocity time graph of a body moving with uniform acceleration uniform retardation. .

107. Draw velocity time graph of a train starting from Ambala and stopping at Delhi, under ideal conditions.

108. Using velocity time graph of uniformly accelerated motion along a straight line, derive the equation for position-velocity relation,

109. A body is moving along a straight line with a uniform acceleration. Draw velocity time graph of this body and obtain the equation for velocity time relation.

110. A passenger lift starts from rest accelerates uniformly and attains a speed of 4-6 m/s in 2 seconds. This speed is maintained for next 6 seconds. Then, the lift undergoes uniform retardation and stops in the next 2 seconds. Draw velocity time graph of the lift and calculate the total distance covered by the lift in 12 seconds.

111. A scooter acquires a velocity of 36 km/h in 10 second after start, and it takes 20 seconds to stop. Calculate acceleration and distance travelled in the two cases.

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112. Draw position time graph of a body
(0 at rest
(«) in uniform motion
(Hi) in uniformly accelerated motion. Explain the curves obtained.

113. Draw speed time graph of a body
(0 at rest
(it) in uniform motion
(HI) in uniformly accelerated motion
(iv) in uniformly retarded motion.
Explain the curves obtained.

114. Explain what is meant by uniform circular motion. Give atleast two examples of this kind of motion.