## Motion Class 9 Physics Worksheet Pdf

Class 9 Physics students should refer to the following printable worksheet in Pdf for Motion in standard 9. This test paper with questions and answers for Grade 9 Physics will be very useful for exams and help you to score good marks

### Class 9 Physics Worksheet for Motion

**CBSE Class 9 Physics Worksheet - Motion**** - Practice worksheets for CBSE students. Prepared by teachers of the best CBSE schools in India.**

**ANSWER THE FOLLOWING QUESTIONS :-**

**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 = t

_{1}

Time taken to reach home = t

_{2}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.

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