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

1.  5 s
2.  10 s
3.  15 s
4.  20 s

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

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

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

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

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

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

(1) $\sqrt{2t}$

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

(3) $\frac{t}{\sqrt{2}}$

(4) Cannot be predicted unless acceleration is given

A particle is moving in a straight line and passes through a point O with a velocity of 6 ms^–1. The particle moves with a constant retardation of 2 ms^–2 for 4 s and there after moves with constant velocity. How long after leaving O does the particle return to O

(1) 3s

(2) 8s

(3) Never

(4) 4s

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., $a=\alpha x^2.$ The distance at which the particle stops is :

1.  $\sqrt{\frac{3v_0}{2\alpha }}$
2.  ${\left(\frac{{\displaystyle 3v_o}}{{\displaystyle 2\alpha }}\right)}^{\frac{1}{3}}$
3.  $\sqrt{\frac{3v{}_0{}^2}{2\alpha }}$
4.  ${\left(\frac{{\displaystyle 3v{}_0{}^2}}{{\displaystyle 2\alpha }}\right)}^{\frac{1}{3}}$  

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

(1) $\overrightarrow{v}/2$ and v/2

(2) 0 and v/2

(3) 0 and 0

(4) $\overrightarrow{v}/2$ and 0

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

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

2. $\sqrt{8g}$ $h$

3. $\sqrt{2g}$ $h$

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

Four marbles are dropped from the top of a tower one after the other at a one-second interval. The first one reaches the ground after 4 seconds. When the first one reaches the ground the distances between the first and second, the second and third, and the third and fourth will be, respectively:

A balloon rises from rest with a constant acceleration g/8. A stone is released from it when it has risen to height h. The time taken by the stone to reach the ground is

(1) $4\sqrt{\frac{h}{g}}$

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

(3) $\sqrt{\frac{2h}{g}}$

(4) $\sqrt{\frac{g}{h}}$ 

Two bodies are thrown simultaneously from a tower with same initial velocity v₀ : one vertically upwards, the other vertically downwards. The distance between the two bodies after time t is

(1) $2v_{0}t+\frac{1}{2}g{t}^2$

(2) 2v₀t

(3) $v_{0}t+\frac{1}{2}g{t}^2$

(4) v₀t