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A thief is running away on a straight road in jeep moving with a speed of 9 ms^–1. A police man chases him on a motor cycle moving at a speed of 10 ms^–1. If the instantaneous separation of the jeep from the motorcycle is 100 m, how long will it take for the police to catch the thief

1 s

19 s

90 s

100 s

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Explanation

The relative velocity of policeman w.r.t. thief = 10 – 9 = 1 m/s.

∴ Time taken by police to catch the thief $= \frac{100}{1}$ =100 sec

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A car A is travelling on a straight level road with a uniform speed of 60 km/h. It is followed by another car B which is moving with a speed of 70 km/h. When the distance between them is 2.5 km, the car B is given a deceleration of 20 km/h². After how much time will B catch up with A

1 hr

1/2 hr

1/4 hr

1/8 hr

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Explanation

Let car B catches, car A after ‘t’ sec, then

$60 t + 2 .5 = 70 t - \frac{1}{2} \times 20 \times t^{2}$

⇒ $10 t^{2} - 10 t + 2 .5 = 0$

⇒ $t^{2} - t + 0 .25 = 0$

∴ $t = \frac{1 \pm \sqrt{1 - 4 \times (0 .25)}}{2} = \frac{1}{2} h r$

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The speed of a body moving with uniform acceleration is u. This speed is doubled while covering a distance S. When it covers an additional distance S, its speed would become

$\sqrt{3} u$

$\sqrt{5} u$

$\sqrt{11} u$

$\sqrt{7} u$

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Explanation

As $v^{2} = u^{2} + 2 a s$ ⇒ ${(2 u)}^{2} = u^{2} + 2 a s \Rightarrow 2 a s = 3 u^{2}$

Now, after covering an additional distance s, if velocity becomes v, then,

$v^{2} = u^{2} + 2 a (2 s) = u^{2} + 4 a s = u^{2} + 6 u^{2} = 7 u^{2}$

∴ $v = \sqrt{7} u$ .

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Two trains one of length 100 m and another of length 125 m, are moving in mutually opposite directions along parallel lines, meet each other, each with speed 10 m/s. If their acceleration are 0.3 m/s² and 0.2 m/s² respectively, then the time they take to pass each other will be

5 s

10 s

15 s

20 s

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Explanation

Relative velocity of one train w.r.t. other

= 10 + 10 = 20 m/s.

Relative acceleration =0.3+0.2=0.5 m/s²

If trains cross each other then from $s = u t + \frac{1}{2} a t^{2}$

$A s, s = s_{1} + s_{2} = 100 + 125 = 225$

⇒ $225 = 20 t + \frac{1}{2} \times 0 .5 \times t^{2}$ ⇒ $0 .5 t^{2} + 40 t - 450 = 0$

⇒ $t = - \frac{40 \pm \sqrt{1600 + 4 . (005) \times 450}}{1} = - 40 \pm 50$

∴ t = 10 sec (Taking +ve value).

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A body starts from rest with uniform acceleration. If its velocity after n second is v, then its displacement in the last two seconds is

$\frac{2 v (n + 1)}{n}$

$\frac{v (n + 1)}{n}$

$\frac{v (n - 1)}{n}$

$\frac{2 v (n - 1)}{n}$

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Explanation

$∵ v = 0 + n a \Rightarrow a = v / n$

Now, distance travelled in n sec. ⇒ $S_{n} = \frac{1}{2} a n^{2}$ and distance travelled in (n – 2) sec ⇒ $S_{n - 2} = \frac{1}{2} a {(n - 2)}^{2}$

∴ Distance travelled in last two seconds,

$= S_{n} - S_{n - 2} = \frac{1}{2} a n^{2} - \frac{1}{2} a {(n - 2)}^{2}$

$= \frac{a}{2} [n^{2} - {(n - 2)}^{2}] = \frac{a}{2} [n + (n - 2)] [n - (n - 2)]$

= $a (2 n - 2) = \frac{v}{n} (2 n - 2) = \frac{2 v (n - 1)}{n}$

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A point starts moving in a straight line with a certain acceleration. At a time t after beginning of motion the acceleration suddenly becomes retardation of the same value. The time in which the point returns to the initial point is

$\sqrt{2 t}$

$(2 + \sqrt{2}) t$

$\frac{t}{\sqrt{2}}$

Cannot be predicted unless acceleration is given

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Explanation

If a particle starts moving with an acceleration and then experiences an equal retardation, the time taken to return to the initial point is (2 + âˆš2) times the time taken to reach the highest point. This can be derived using kinematic equations and the symmetry of the motion.

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A bird flies for 4 s with a velocity of $| t - 2 | m / s$ in a straight line, where t is time in seconds. It covers a distance of

2 m

4 m

6 m

8 m

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A particle is projected with velocity $v_{0}$ along x-axis. The deceleration of the particle is proportional to the square of the distance from the origin i.e., $a = α x^{2} .$ The distance at which the particle stops is :

$\sqrt{\frac{3 v_{0}}{2 α}}$

${(\frac{3 v_{o}}{2 α})}^{\frac{1}{3}}$

$\sqrt{\frac{3 v {}_{0}^{2}}{2 α}}$

${(\frac{3 v {}_{0}^{2}}{2 α})}^{\frac{1}{3}}$

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Explanation

$a = \frac{d v}{d t} = \frac{d v}{d x} \frac{d x}{d t} = v \frac{d v}{d x} = - α x^{2}$ (given)

⇒ $\int_{v_{0}}^{0} v d v = - α \int_{0}^{S} x^{2} d x$ ⇒ ${[\frac{v^{2}}{2}]}_{v_{0}}^{0} = - α {[\frac{x^{3}}{3}]}_{0}^{S}$

⇒ $\frac{v_{0}^{2}}{2} = \frac{α S^{3}}{3}$ ⇒ $S = {[\frac{3 v_{0}^{2}}{2 α}]}^{\frac{1}{3}}$

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A body is projected vertically up with a velocity v and after some time it returns to the point from which it was projected. The average velocity and average speed of the body for the total time of flight are

$\vec{v} / 2$ and v/2

0 and v/2

0 and 0

$\vec{v} / 2$ and 0

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Explanation

Average velocity = 0 because net displacement of the body is zero.

Average speed $= \frac{Total distance covered}{Time of flight} = \frac{2 H_{\max}}{2 u / g}$

⇒ $v_{a v} = \frac{2 u^{2} / 2 g}{2 u / g}$ ⇒ $v_{a v} = u / 2$

Velocity of projection = v (given)

∴ $v_{av} = v / 2$

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A stone is dropped from a height h. Simultaneously, another stone is thrown up from the ground which reaches a height 4 h. The two stones cross each other after time

$\sqrt{\frac{h}{8 g}}$

$\sqrt{8 g h}$

$\sqrt{2 g h}$

$\sqrt{\frac{h}{2 g}}$

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