Q117 / 150 · MHT CET 2023 (13 May Shift 2) · Physics

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

A
$\frac{2 \mathrm{~V}}{\mathrm{a}}$
B
$\frac{a}{2 V}$
Correct answer
C
$\frac{\mathrm{V}}{2 \mathrm{a}}$
D
$\sqrt{\frac{V}{a}}$

Solution

At point $A$, $\mathrm{S}_{\mathrm{A}}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2$ $\mathrm{S}_{\mathrm{A}}=\frac{1}{2} \mathrm{at}^2$ At point $B$, $\mathrm{S}_{\mathrm{B}}=\mathrm{vt}$ $\therefore \quad$ At the point where they meet, $\mathrm{S}_{\mathrm{A}}=\mathrm{S}_{\mathrm{B}}$ $\frac{1}{2} a t^2=v t$ $t=\frac{2 v}{a}$

Hence, the correct answer is option: 'C'.

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