Question

A particle starts moving along a line from rest and comes to rest after moving distance $d$. During its motion, it had a constant acceleration $f$ over $2/3$ of the distance and covered the rest of the distance with constant retardation. The time taken to cover the distance is:

• A. $\sqrt{\frac{2d}{3f}}$
• B. $2\sqrt{\frac{d}{3f}}$
• C. $\sqrt{\frac{3d}{f}}$
• D. $\sqrt{\frac{3d}{2f}}$

Answer 1

Answer: (C)

$\sqrt{\frac{3d}{f}}$

Let velocity at B be $v$.

From A to B :

Initial velocity, $u=0$

Acceleration, $a=f$

Time taken, $t=t_1$

Using Newton's first equation of motion,

$v=u+at$

Velocity at B, $v_1=0+f{t}_1=f{t}_1$

$v_1=f{t}_1$ ....(1)

Using Newton's second equation of motion,

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

Distance covered, $s=\frac{2d}{3}$

$\frac{2d}{3}=0+\frac{1}{2}f{t}_1^2$

$⟹t_1=\sqrt{\frac{4d}{3f}}$

$⟹t_1=2\sqrt{\frac{d}{3f}}$ .....(b)

From B to C :

Initial velocity, $u=v_1$

Final velocity, $v=0$

Acceleration, $a$

Time taken, $t=t_2$

Using Newton's first equation of motion,

$v=u+at$

$0=v_1+a{t}_2$

$⟹=v=-a{t}_2$ ....(2)

From (1) and (2), we get

$a=-\frac{f{t}_1}{t_2}$

Using Newton's third equation of motion

$2as=v^2-u^2$

$2a\frac{d}{3}=0^2-v_1^2$

$\frac{d}{3}=\frac{-v^2}{2a}=\frac{-(f{t}_1{)}^2}{2\times (-f{t}_1/t_2)}=\frac{f{t}_1{t}_2}{2}$ ....(a)

From (a) and (b), $\frac{d}{3}=\frac{f{t}_2}{2}\times \sqrt{\frac{4d}{3f}}$

$⟹t_2=\sqrt{\frac{d}{3f}}$

Thus total time taken, $t=t_1+t_2=2\sqrt{\frac{d}{3f}}+\sqrt{\frac{d}{3f}}=3\sqrt{\frac{d}{3f}}$

$t=\sqrt{\frac{3d}{f}}$