SATHEE: Chapter 07 Motion Questions-05

Chapter 07 Motion Questions-05

Questions

1. 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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Answer

Initial speed of the bus, $u = 0$

Acceleration, $a = 0.1 m / s^{2}$

Time taken, $t = 2$ minutes $= 120 s$

(a) $v = u + a t$

$v = 0 + 0 \times 1 \times 120$

$v = 12 m s^{- 1}$

(b) According to the third equation of motion:

$v^{2} - u^{2} = 2 a s$

Where, $s$ is the distance covered by the bus

$(12)^{2} - (0)^{2} = 2 (0.1) s$

$s = 720 m$

Speed acquired by the bus is $12 m / s$ .

Distance travelled by the bus is $720 m$ .

2. A train is travelling at a speed of $90 k m h^{- 1}$ . 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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Answer

Initial speed of the train, $u = 90 k m / h = 25 m / s$

Final speed of the train, $v = 0$ (finally the train comes to rest)

Acceleration $= - 0.5 m s^{- 2}$

According to third equation of motion:

$v^{2} = u^{2} + 2$ as

$(0)^{2} = (25)^{2} + 2 (- 0.5) s$

Where, $s$ is the distance covered by the train

$s = \frac{25^{2}}{2 (0.5)} = 625 m$

The train will cover a distance of $625 m$ before it comes to rest.

3. A trolley, while going down an inclined plane, has an acceleration of $2 c m s^{- 2}$ . What will be its velocity $3 s$ after the start?

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Answer

Initial Velocity of trolley, $u = 0 c m s^{- 1}$

Acceleration, $a = 2 c m s^{- 2}$

Time, $t = 3 s$

We know that final velocity, $v = u + a t = 0 + 2 \times 3 c m s^{- 1}$

Therefore, The velocity of train after 3 seconds $= 6 c m s^{- 1}$

4. 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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Answer

Initial Velocity of the car, $u = 0 m s^{- 1}$

Acceleration, $a = 4 m s^{- 2}$

Time, $t = 10 s$

We know Distance, $s = u t + (1 / 2) a t^{2}$

Therefore, Distance covered by car in 10 second $= 0 \times 10 + (1 / 2) \times 4 \times$ 102

$= 0 + (1 / 2) \times 4 \times 10 \times 10 m$

$= (1 / 2) \times 400 m$

$= 200 m$

5. 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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Answer

Given Initial velocity of stone, $u = 5 m s^{- 1}$

Downward of negative Acceleration, $a = 10 m s^{- 2}$

We know that $2 a s = v^{2} - u^{2}$

Therefore, Height attained by the stone, $s = \frac{0^{2}}{5^{2}} \times (- 10) m$

$= \frac{- 25}{- 20} m$

$= 1.25 m$

Also we know that final velocity, $v = u + a t$

or, Time, $t = \frac{v - u}{a}$

Therefore, Time, $t$ taken by stone to attain the height, $s = \frac{0 - 5}{- 10 s}$ $= 0.5 s$

Chapter 07 Motion Questions-05

Questions

Table of Contents

Engineering Preparation

Medical Preparation

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NCERT Exercise Video Solution

Chapter 07 Motion Questions-05

Questions

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Mentorship

NCERT Exercise Video Solution

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