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A boggy of uniformly moving train is suddenly detached from train and stops after covering some distance. The distance covered by the boggy and distance covered by the train in the same time has relation

Both will be equal

First will be half of second

First will be 1/4 of second

No definite ratio

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Explanation

Let 'a' be the retardation of boggy then distance covered by it be S. If u is the initial velocity of boggy after detaching from train (i.e. uniform speed of train)

$v^{2} = u^{2} + 2 a s \Rightarrow 0 = u^{2} - 2 a s \Rightarrow s_{b} = \frac{u^{2}}{2 a}$

Time taken by boggy to stop

$v = u + a t \Rightarrow 0 = u - a t \Rightarrow t = \frac{u}{a}$

In this time t distance travelled by train $= s_{t} = u t = \frac{u^{2}}{a}$

Hence ratio $\frac{s_{b}}{s_{t}} = \frac{1}{2}$

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A body starts from rest. What is the ratio of the distance travelled by the body during the 4th and 3rd second

$\frac{7}{5}$

$\frac{5}{7}$

$\frac{7}{3}$

$\frac{3}{7}$

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Explanation

$S_{n} = u + \frac{a}{2} (2 n - 1) = \frac{a}{2} (2 n - 1)$ because $u = 0$

Hence $\frac{S_{4}}{S_{3}} = \frac{7}{5}$

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The acceleration ‘a’ in m/s² of a particle is given by $a = 3 t^{2} + 2 t + 2$ where t is the time. If the particle starts out with a velocity u = 2 m/s at t = 0, then the velocity at the end of 2 second is

12 m/s

18 m/s

27 m/s

36 m/s

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Explanation

$v = u + \int_{}^{} a d t = u + \int_{}^{} (3 t^{2} + 2 t + 2) d t$

$= u + \frac{3 t^{3}}{3} + \frac{2 t^{2}}{2} + 2 t = u + t^{3} + t^{2} + 2 t$

$= 2 + 8 + 4 + 4 = 18 m / s$ (As t = 2 sec)

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A particle moves along a straight line such that its displacement at any time t is given by $S = t^{3} - 6 t^{2} + 3 t + 4$ metres

The velocity when the acceleration is zero is

4 $m s^{- 1}$

$- 12 m s^{- 1}$

$42 m s^{- 1}$

$- 9 m s^{- 1}$

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Explanation

$v = \frac{d s}{d t} = 3 t^{2} - 12 t + 3$ and $a = \frac{d v}{d t} = 6 t - 12$

For $a = 0$ , we have $t = 2$ and at $t = 2, v = - 9 m s^{- 1}$

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If a body starts from rest and travels 120 cm in the 6^th second, then what is the acceleration

0.20 m/s²

0.027 m/s²

0.218 m/s²

0.03 m/s²

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Explanation

$S_{n} = u + \frac{a}{2} (2 n - 1)$ ⇒ $1 .2 = 0 + \frac{a}{2} (2 \times 6 - 1)$

⇒ $a = \frac{1 .2 \times 2}{11} = 0 .218 m / s^{2}$

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If a car at rest accelerates uniformly to a speed of 144 km/h in 20 s. Then it covers a distance of

20 m

400 m

1440 m

2880 m

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Explanation

Here $v = 144 k m / h = 40 m / s$

$v = u + a t$ ⇒ $40 = 0 + 20 \times a \Rightarrow a = 2 m / s^{2}$

∴ $s = \frac{1}{2} a t^{2} = \frac{1}{2} \times 2 \times {(20)}^{2} = 400 m$

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The position x of a particle varies with time t as $x = a t^{2} - b t^{3}$ . The acceleration of the particle will be zero at time t equal to

$\frac{a}{b}$

$\frac{2 a}{3 b}$

$\frac{a}{3 b}$

Zero

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Explanation

$\frac{d x}{d t} = 2 a t - 3 b t^{2} \Rightarrow \frac{d^{2} x}{d t^{2}} = 2 a - 6 b t = 0 \Rightarrow t = \frac{a}{3 b}$

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If a train travelling at 72 kmph is to be brought to rest in a distance of 200 metres, then its retardation should be

20 ms^–2

10 ms^–2

2 ms^–2

1 ms^–2

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Explanation

$u = 72 k m p h = 20 m / s,$

By using $v^{2} = u^{2} - 2 a s$ ⇒ $a = \frac{u^{2}}{2 s} = \frac{{(20)}^{2}}{2 \times 200} = 1 m / s^{2}$

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The displacement of a particle starting from rest (at t = 0) is given by $s = 6 t^{2} - t^{3}$ . The time in seconds at which the particle will attain zero velocity again, is

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Explanation

$v = \frac{d s}{d t} = 12 t - 3 t^{2}$

Velocity is zero for $t = 0$ and $t = 4 \sec$

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Two cars A and B are at rest at same point initially. If A starts with uniform velocity of 40 m/sec and B starts in the same direction with constant acceleration of 4 m/s², then B will catch A after how much time ?

10 sec

20 sec

30 sec

35 sec

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Explanation

Let A and B will meet after time t sec. it means the distance travelled by both will be equal.

$S_{A} = u t = 40 t$ and $S_{B} = \frac{1}{2} a t^{2} = \frac{1}{2} \times 4 \times t^{2}$

$S_{A} = S_{B} \Rightarrow 40 t = \frac{1}{2} 4 t^{2} \Rightarrow t = 20$ sec

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