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A car, starting from rest, accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate $\frac{f}{2}$ to come to rest. If the total distance traversed is 15 S, then

$S = \frac{1}{2} f t^{2}$

$S = \frac{1}{4} f t^{2}$

$S = \frac{1}{72} f t^{2}$

$S = \frac{1}{6} f t^{2}$

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A man is 45 m behind the bus when the bus starts accelerating from rest with acceleration of 2.5 m/s². With what minimum velocity should the man start running to catch the bus?

12 m/s

14 m/s

15 m/s

16 m/s

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Explanation

Let man will catch the bus after 't' sec . So he will cover distance ut.

Similarly distance travelled by the bus will be $\frac{1}{2} a t^{2}$ . For the given condition

$u t = 45 + \frac{1}{2} a t^{2} = 45 + 1 .25 t^{2}$ $[A s a = 2 .5 m / s^{2}]$

⇒ $u = \frac{45}{t} + 1 .25 t$

To find the minimum value of u

$\frac{d u}{d t} = 0$

so we get $t = 6 \sec$ then,

$u = \frac{45}{6} + 1 .25 \times 6 = 7 .5 + 7 .5 = 15 m / s$

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A particle moves along x-axis as $x = 4 (t - 2) + a {(t - 2)}^{2}$ . Which of the following is true ?

The initial velocity of particle is 4

The acceleration of particle is 2a

The particle is at origin at t = 0

None of these

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Explanation

$x = 4 (t - 2) + a {(t - 2)}^{2}$

At $t = 0, x = - 8 + 4 a = 4 a - 8$

$v = \frac{d x}{d t} = 4 + 2 a (t - 2)$

At $t = 0, v = 4 - 4 a = 4 (1 - a)$

But acceleration, $a = \frac{d^{2} x}{d t^{2}} = 2 a$

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A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the 5th sec to that covered in 5 sec is

9/25

3/5

25/9

1/25

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Explanation

Distance covered in 5^th second,

$S_{5^{t h}} = u + \frac{a}{2} (2 n - 1) = 0 + \frac{a}{2} (2 \times 5 - 1) = \frac{9 a}{2}$

and distance covered in 5 second,

$S_{5} = u t + \frac{1}{2} a t^{2} = 0 + \frac{1}{2} \times a \times 25 = \frac{25 a}{2}$

∴ $\frac{S_{5^{t h}}}{S_{5}} = \frac{9}{25}$

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What determines the nature of the path followed by the particle

Speed

Velocity

Acceleration

Both (2) and (3)

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Explanation

The nature of the path is decided by the direction of velocity, and the direction of acceleration. The trajectory can be a straight line, circle or a parabola depending on these factors.

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Two trains, each 50 m long are travelling in opposite direction with velocity 10 m/s and 15 m/s. The time of crossing is

10 s

4 s

$2 \sqrt{3} s$

$4 \sqrt{3} s$

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Explanation

Time $= \frac{Total length}{Relative velocity} = \frac{50 + 50}{10 + 15} = \frac{100}{25} = 4 s e c$

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A 120 m long train is moving in a direction with speed 20 m/s. A train B moving with 30 m/s in the opposite direction and 130 m long crosses the first train in a time

4 s

36 s

38 s

5 s

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Explanation

Total distance $= 130 + 120 = 250 m$

Relative velocity $= 30 - (- 20) = 50 m / s$

Hence $t = 250 / 50 = 5 s$

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A 210 meter long train is moving due North at a speed of 25m/s. A small bird is flying due South a little above the train with speed 5m/s. The time taken by the bird to cross the train is

6 s

7 s

9 s

10 s

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Explanation

Relative velocity of bird w.r.t train $= 25 + 5 = 30 m / s$

time taken by the bird to cross the train $t = \frac{210}{30} = 7 s e c$

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A police jeep is chasing with, velocity of 45 km/h a thief in another jeep moving with velocity 153 km/h. Police fires a bullet with muzzle velocity of 180 m/s. The velocity it will strike the car of the thief is

150 m/s

27 m/s

450 m/s

250 m/s

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Explanation

Effective speed of the bullet

= speed of bullet + speed of police jeep

$= 180 m / s + 45 k m / h = (180 + 12 .5) m / s = 192 .5 m / s$

Speed of thief ’s jeep $= 153 k m / h = 42 .5 m / s$

Velocity of bullet w.r.t thief ’s car $= 192 .5 - 42 .5$ = 150m/s

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A train of 150 meter length is going towards north direction at a speed of 10 m/sec. A parrot flies at the speed of 5 m/sec towards south direction parallel to the railway track. The time taken by the parrot to cross the train is

12 sec

8 sec

15 sec

10 sec

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Explanation

Relative velocity

$= 10 + 5 = 15 m / s e c$

∴ $t = \frac{150}{15} = 10 s e c$

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