Class 9th Science - Motion Case Study Questions and Answers 2022 - 2023
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Motion Case Study Questions With Answer Key
9th Standard CBSE
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Reg.No. :
Science
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Suppose the boy first runs a distance of 100 metres in 50 seconds in going from his home to the shop in the East direction, and then runs a distance of 100 metres again. in 50 seconds in the reverse direction from the shop to reach back home from where he started (see Figure).
(i) Find the speed of the boy.(a) 1 m/s (b) 2 m/s (c) 3 m/s (d) none of these (ii) Find the Velocity of the boy.
(a) 1 m/s (b) 2 m/s (c) 3 m/s (d) 0 m/s (iii) A boy is sitting on a merry-go-round which is moving with a constant speed of 10m/s. This means that the boy is :
(a) at rest (b) moving with no acceleration (c) in accelerated motion (d) moving with uniform velocity (iv) In which of the following cases of motion, 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 on circular road (c) if the pendulum is moving to and fro (d) if a planet is moving around the sun (v) A particle is moving in a circular path of radius r. The displacement after half a circle would be :
(a) 0 (b) πr (c) 2r (d) 2πr -
One day Rahul decided to go his office by his car. He is enjoying the driving along with listening the old songs. His car is moving along a straight road at a steady speed. On a particular moment, he notices that the car travels 150 m in 5 seconds.
(i) What is its average speed ?(a) 20 m/s (b) 30 m/s (c) 10 m/s (d) 40 m/s (ii) How far does it travel in 1 second ?
(a) 20 m (b) 30 m (c) 10 m (d) 40 m (iii) How far does it travel in 6 seconds ?
(a) 120 m (b) 130 m (c) 180 m (d) 140 m (iv) How long does it take to travel 240 m ?
(a) 2s (b) 4s (c) 6s (d) 8s (v) Which of the following statement is correct regarding velocity and speed of a moving body?
(a) velocity of a moving body is always higher than its speed
(b) speed of a moving body is always higher than its velocity
(c) speed of a moving body is its velocity in a given direction
(d) velocity of a moving body is its speed in a given direction -
Aditya started driving his car. He increases the speed till 4 seconds and then he kept his card in constant speed for 6 seconds. Then after he decreased the speed of the car upto another 6 seconds. After reaching at the starting place, he draws the speed-time graph of his 16 seconds driving as shown below:
(i) What type of motion is represented by OA ?(a) uniform velocity (b) uniform acceleration (c) negative acceleration (d) no acceleration (ii) What type of motion is represented by BC ?
(a) uniform velocity (b) uniform acceleration (c) negative acceleration (d) no acceleration (iii) Find out the acceleration of the body.
(a) 1.5 m/s2 (b) 2 m/s2 (c) 3 m/s2 (d) 1 m/s2 (iv) Calculate the retardation of the body.
(a) 1.5 m/s (b) 2 m/s2 (c) 3 m/s2 (d) 1 m/s2 (v) Find out the distance travelled by the body from A to B.
(a) 15 m (b) 30 m (c) 36 m (d) 60 m
Case Study
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Answers
Motion Case Study Questions With Answer Key Answer Keys
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(i) (b) 2 m/s
Total distance travelled is 100 m + 100 m = 200 m and the total time taken is 50 s + 50 s = 100 s.
\(\text { Speed of boy }=\frac{\text { Distance travelled }}{\text { Time taken }}=\frac{200 \mathrm{~m}}{100 \mathrm{~s}}=2 \mathrm{~m} / \mathrm{s}\)
(ii) (d) 0 m/s
The boy runs 100 m towards East and then 100 m towards West and reaches at the starting point, his home. So, the displacement will be 100 m – 100 m = 0 m.
The total time taken is 50 s + 50 s = 100 s.
\(\text { Velocity of boy }=\frac{\text { Displacement }}{\text { Time taken }}=\frac{0 \mathrm{~m}}{100 \mathrm{~s}}=0 \mathrm{~m} / \mathrm{s}\)
(iii) (c) in accelerated motion
(iv) (a) if the car is moving on straight road
(v) (c) 2r -
(i) (b) 30 m/s
Average speed = total distance travelled/total time taken
= 150/5
= 30 m/s
(ii) (b) 30 m
Time = 1 s
Distance = (average speed)(time)
= 30 m/s x 1s
= 30 m
(iii) (c) 180 m
Time = 6 s
Distance = (average speed)(time)
= 30 m/s x 6s
= 180m
(iv) (d) 8s
Distance = 240m
Time = Distance/average speed
= 240/30
= 8s
(v) (d) velocity of a moving body is its speed in a given direction -
(i) (b) uniform acceleration
OA is a straight line graph between speed and time, and it is sloping upwards from O to A.
Therefore, the graph line OA represents uniform acceleration.
(ii) (c) negative acceleration
BC is a straight line graph between speed and time which is sloping downwards from B to C.
Therefore, BC represents uniform retardation (or negative acceleration).
(iii) (a) 1.5 m/s2
The slope of speed-time graph OA will give us the acceleration of the body.
Thus, Acceleration = Slope of line OA
= AD/OD
Now, in the given graph, we find that AD = 6 m/s and OD
= 4 seconds.
So, putting these values in the above relation, we get :
Acceleration = 6 m/s / 4 s
= 1.5 m/s2
(iv) (d) 1 m/s2
The slope of speed-time graph BC will be equal to the retardation of the body.
So, Retardation = Slope of line BC
= BE/EC
Now, in the graph given to us, we find that BE = 6 m/s and EC = 16 – 10 = 6 seconds.
So, putting these values in the above relation, we get :
Retardation = 6m/s / 6 s
= 1 m/s2
(v) (c) 36 m
Distance travelled from A to B
= Area under the line AB and the time axis
= Area of rectangle DABE
= DA × DE
Now, from the given graph, we find that DA = 6 m/s and DE = 10 – 4 = 6 s.
Therefore, Distance travelled from A to B = 6 × 6
= 36 m
