A body moves with uniform acceleration a and zero initial velocity. another body B starts from the same point and moves in same direction with constant velocity v. the two bodies meet after a time t. the value of t is ?

Question

options:-
1. 2v/a
2. v/a
3. v/2a
4. $\sqrt{\frac{v}{2 a}}$ $sqrt(v/(2a))$

25818 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-05T13:55:20

The answer is option $(1)$ $(1)$ , i.e $= \frac{2 v}{a}$ $=(2v)/a$

We apply the equation of motion to the body $A$ $A$

$s = u t + \frac{1}{2} a t^{2}$ $s=ut+1/2at^2$

The initial velocity of $A$ $A$ is $u_{A} = 0$ $u_A=0$

The aceleration is $a_{A} = a$ $a_A=a$

So, the distance after the time $t$ $t$ i

$s = 0 + \frac{1}{2} a t^{2}$ $s=0+1/2at^2$

We apply the equation of motion to the body $B$ $B$

$s = v t$ $s=vt$

Therefore,

$v t = \frac{1}{2} a t^{2}$ $vt=1/2at^2$

$t = \frac{2 v}{a}$ $t=(2v)/a$