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NCERT · Class 9 · Science

Chapter 7: Motion

Practice each question, then open the answer or worked solution.

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Intext Questions – 5

Textbook page 109,110 · 5 questions
Q1Written answer

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.

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Worked solution

(a) Given, the bus 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 a=(v-u)/t

Therefore, terminal velocity (v) = (at)+u

= (0.1 m.s-2 * 120 s) + 0 m.s-1

= 12 m.s-1 + 0 m.s-1

Therefore, terminal velocity (v) = 12 m/s

(b) As per the third motion equation, 2as = v2 – u2

Since a = 0.1 m.s-2, v = 12 m.s-1, u = 0 m.s-1, and t = 120 s, the following value for s (distance) can be obtained.

Distance, s =(v2 – u2)/2a

=(122 – 02)/2(0.1)

Therefore, s = 720 m.

The speed acquired is 12 m.s-1 and the total distance travelled is 720 m.

Q2Written answer

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.

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Worked solution

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, v2-u2=2as

Therefore, distance traveled by the train (s) =(v2-u2)/2a

s = (02-252)/2(-0.5) meters = 625 meters

The train must travel 625 meters at an acceleration of -0.5 ms-2 before it reaches the rest position.

Q3Written answer

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?

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Worked solution

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, v=u+at

Therefore, terminal velocity of the trolley (v) = 0 + (0.02 ms-2)(3s)= 0.06 ms-1

Therefore, the velocity of the trolley after 3 seconds will be 6 cm.s-1

Q4Written answer

A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start?

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Worked solution

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, s = ut+1/2 at2

Therefore, the total distance covered by the car (s) = 0 * 10m + 1/2 (4ms-2)(10s)2

= 200 meters

Therefore, the car will cover a distance of 200 meters after 10 seconds.

Q5Written answer

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?

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Worked solution

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, v2 – u2 = 2as

Therefore, the distance travelled by the stone (s) = (02 – 52)/ 2(10)

Distance (s) = 1.25 meters

As per the first motion equation, v = u + at

Therefore, time taken by the stone to reach a position of rest (maximum height) = (v – u) /a

=(0-5)/-10 s

Time taken = 0.5 seconds

Therefore, the stone reaches a maximum height of 1.25 meters in a timeframe of 0.5 seconds.