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

Step 1: Write each distance.
Body B moves steadily: $x_B=Vt$. Body A speeds up from rest: $x_A=\frac{1}{2}at^2$.

Step 2: Set them equal at meeting.
They start together and meet again, so $\frac{1}{2}at^2=Vt$.

Step 3: Cancel $t$ and solve.
Dividing by $t$, $\frac{1}{2}at=V$, so $t=\frac{2V}{a}$. \[ \boxed{\frac{2V}{a}} \]