Question

Two bodies 'A' and 'B' start from the same point at the same instant and move along a straight line. 'A' moves with uniform acceleration ($a$) and 'B' moves with uniform velocity ($V$). They meet after time '$t$'. The value of '$t$' is

• A. $\frac{2V}{a}$
• B. $\frac{V}{a}$
• C. $\frac{a}{2V}$
• D. $\frac{V}{2a}$

Hint:

For a uniformly accelerating object starting from rest, its average velocity over a time interval $t$ is exactly half of its final velocity ($v_{\text{avg}} = \frac{at}{2}$). For it to catch up to an object moving with constant velocity $V$, their average velocities must match: $\frac{at}{2} = V \implies t = \frac{2V}{a}$ instantly!

Answer 1

The Correct Option is A

Let's set up the kinematic displacement equations for both bodies starting from the origin ($x_0 = 0$): * For body B moving with a constant uniform velocity $V$, its displacement after time $t$ is: $$x_B = V \cdot t$$ * For body A starting from rest ($u = 0$) with uniform acceleration $a$, its displacement after time $t$ is: $$x_A = \frac{1}{2}a t^2$$ Since they meet after time $t$, their total displacements must be exactly equal ($x_A = x_B$): $$\frac{1}{2}a t^2 = V t$$ Since $t \neq 0$ at the meeting point, we divide both sides by $t$: $$\frac{1}{2}a t = V \implies a t = 2V \implies t = \frac{2V}{a}$$
Final Answer:
The value of time $t$ when they meet is $\frac{2V}{a}$, which corresponds to option (A).