Motion Class 9 Science Important Questions

Motion Class 9 Science Chapter 8 Important Questions

Hello every one, here in this post you will get the important questions of Chapter 8 - Motion of Class 9 Science. Do practice all these questions from the Motion chapter along with the questions given in your NCERT textbook to get the maximum score in your exam. These questions are compiled by CBSE Guidance.

Q. No. 1) A particle is moving in a circular path of radius (r). The displacement after half a circle would be

a. Zero

b. πr

c. 2r

d. 2πr

Ans. Option c. (2r)

Q. No. 2) An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Ans. Yes. If an object starts its motion from point A and after traveling some distance, it returns back to point A, then the displacement is zero. That is, if the initial position and the final position of an object coincide, then the displacement is zero.

Q. No. 3) The following velocity-time graph shows the motion of a cyclist.

velocity-time graph class 9 numerical

Find

i. Its acceleration

ii. Its velocity

iii. The distance covered by the cyclist in 15 seconds.

Ans. i. a = 0 m/s2, because the v-t graph is a straight line parallel to the time axis, i.e., the body is in uniform motion.

ii. v = 20 m/s

iii. s = area under the v-t graph = 20 x 15 = 300 m.

Q. No. 4) Distinguish between distance and displacement.

Ans.

Distance Displacement
1. It is the total path length covered by the object. 1. It is the shortest distance measured from the initial to the final position of an object.
2. It is independent of direction (a scalar quantity) 2. It is dependent on direction (vector quantity)
3. For a course of motion, the distance covered can never be zero. 3. For a course of motion, displacement can be zero.
4. Distance is always greater than or equal to displacement. 4. Displacement is always smaller than or equal to distance.

Q. No. 5) An object travels 16 m in 4 seconds and then another 16 m in 3 seconds. Is his motion uniform or non-uniform? Why?

Ans. Non-uniform, because the object covers equal distance in unequal intervals of time.

Q. No. 6) The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of the trip. If the trip took 8 h, calculate the average speed of the car in km/h and m/s.

Ans. Distance (s) = 2400 – 2000 = 400 km

Time (t) = 8 hr

Average speed = s/t = 400/8 = 50 km/h = 50 x 5/18 = 13.9 m/s.

Q. No. 7) The numerical ratio of displacement to distance covered by 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

Ans. Option (d)

Also See: Science Class 9 Important Questions

Q. No. 8) Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m/s in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m/s in the next 5 s. Calculate the acceleration of the bicycle in both cases.

Ans. First Case:

u = 0 m/s

v = 6 m/s

t = 30 s

a = (v-u)/t = (6-0)/30 = 6/30 = 0.2 ms^(-2)

Second Case:

u = 6 m/s

v = 4 m/s

t = 5 s

a = (v-u)/t = (4-6)/5 = (-2)/5 = -0.4 ms^(-2)

Q. No. 9) For an object moving with uniform velocity, how will the first equation of motion change?

Ans. For a uniform motion,

a = 0

The first equation of motion:

v = u + at

⇒ v = u + (0)t

⇒ v = u

Q. No. 10) What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

Ans. The object is at rest.

Q. No. 11) A motorcyclist drives from point A to point B with a uniform speed of 30 km/h and returns back to point A with a uniform speed of 20 km/h. Find the average speed of the motorcyclist.

Ans. motion class 9 science A motorcyclist drives from point A to point B with a uniform speed of 30 km/h and returns back to point A with a uniform speed of 20 km/h. Find the average speed of the motorcyclist.

Q. No. 12) What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

Ans. The object is moving with a uniform/constant speed.

Q. No. 13) Why is the motion in a circle at a uniform speed called accelerated motion?

Ans. When a body moves along a circular path with a constant speed, its direction of motion at any point is along the tangent to the circle at that point, this motion is called accelerated motion. The direction of motion changes as the body moves in a circle and causes a change in velocity. Therefore, the motion of an object along a circular path is an accelerated motion.

Q. No. 14) What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

Ans. For uniform motion: Straight line parallel to the time axis

For non-uniform motion: Curved line/Non–linear.

Q. No. 15) Distinguish between speed and velocity.

Ans.

Speed  Velocity
1. It is the rate of change in distance. 1. It is the rate of change of displacement.
2. It is independent of direction (Scalar quantity) 2. It is dependent on direction (Vector quantity)
3. It cannot be zero for a course of motion. 3. It can be zero for a course of motion (if displacement=0).

Q. No. 16) The brakes applied to a car produce an acceleration of 6 ms^(-2) in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time.

Ans. motion class 9 numericals

Q. No. 17) A train moves with a speed of 30 km/h in the first 15 minutes, with another speed of 40 km/h in the next 15 minutes, and then with a speed of 60 km/h in the last 30 minutes. Calculate the average speed of the train for this journey.

Ans.

Distance 1 = 30 x 15/60 = 7.5 km

Distance 2 = 40 x 15/60 = 10 km

Distance 3 = 60 x 30/60 = 30 km

Total distance = 47.5 km

Total time = 15/60 + 15/60 + 30/60 = 1 hour

Average speed = (Total Distance)/(Total Time) = 47.5/1 = 47.5 km/h

Q. No. 18) A cyclist moving on a circular track of a radius of 100 m completes one revolution in 2 minutes. What are the average speed and average velocity?

Ans.

Distance travelled in one revolution = 2πr = 2π x 100 = 200 π

Total time = 2 minutes = 2 x 60 = 120 s

Average speed = (Total Distance)/(Total Time) = (200π)/120 = (5π)/3 m/s

Average velocity = (Total Displacement)/(Total Time) = 0/120 = 0 m/s

Q. No. 19) Define uniform circular motion and give the expression of velocity for an object exhibiting uniform circular motion.

Ans. When an object moves in a circular path with uniform speed, its motion is called uniform circular motion.

v = 2πr/t

Q. No. 20) From the given v-t graph, it can be inferred that the object is

a. in uniform motionvelocity-time graph of an object parallel to time axis

b. at rest

c. in non-uniform motion

d. moving with uniform acceleration 

Ans. Option (a)

Q. No. 21) Area under the v-t graph represents a physical quantity that has the unit

motion class 9 science

Ans. Option (b)

Q. No. 22) Figure shows a distance-time graph of two objects A and B, which object is moving with greater speed when both are moving?Distance-time graph question

Ans. Object B is moving with greater speed.

Q. No. 23) The average speed of a bicycle, an athlete, and a car are 18 km/h, 7 m/s, and 2 km/min respectively. Which of the three is the fastest and which is the slowest?

Ans. 18 km/h = 18 x 5/18 = 5 m/s

2 km/ min = (2 km)/(1 min) = (2000 m)/(60 s) = 33.3 m/s

The average speed of the bicycle, the athlete, and the car are 5 m/s, 7 m/s, and 33.3 m/s respectively.

So, the car is the fastest and the bicycle is the slowest.

Q. No. 24) The distance-time graph of the two trains is given below. The trains start simultaneously in the same direction.

i. How much ahead of A is B when the motion starts?

ii. What is the speed of B?

iii. When and where will A catch B?

iv. What is the difference between the speeds of A and B?

v. Is the speed of both the trains uniform or non-uniform? Justify your answer. 

Distance-time graph of two objects A and B

Ans.

i. 100 km

ii. Speed of B = (S2-S1)/(t2 -t1) = (150-100)/(2-0) = 50/2 = 25 km/h

iii. A will catch B at point Q after 2 hours and at a distance of 150 km.

iv. Speed of A = (S2-S1)/(t2 -t1) = (150-0)/(2-0) = 150/2 = 75 km/h

Therefore, Difference between speed of A and B = 75 km/h – 25 km/h = 50 km/h

v. Speed of both the trains is Uniform because both are straight line graphs.

Recommended: Also Watch the Full Explanation of this chapter here:

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2 thoughts on “Motion Class 9 Science Important Questions”

  1. Amazing content sir..thanx a lot… KINDKY complete all chapters for ninth on YouTube and upload the important questions on the website as well…. Very helpful. Thanx once again

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