Question

A particle moves with uniform acceleration along a straight line $AB$. Its velocities at $A$ and $B$ are $2m/s$ and $14m/s$, respectively. $M$ is the mid-point of $AB$. The particle takes $t_1$ seconds to go from $A$ to $M$ and $t_2$ seconds to go from $M$ to $B$. Then $t_2/t_1$ is

• A. $1:1$
• B. $2:1$
• C. $1:2$
• D. $3:1$

Answer 1

Answer: (C)

$v^2=2^2+\frac{2as}{2}$

$\Rightarrow v^2-4=as\dots (i)$
$14^2-v^2=\frac{2as}{2}$
$\Rightarrow 196-v^2=as\dots (ii)$
From (i) and (ii),
$v^2-4=196-v^2$
$\Rightarrow v=10m/s$
Now $t_1=\frac{v-2}{a}=\frac{10-2}{a}=\frac{8}{a}$
$t_2=\frac{14-v}{a}=\frac{14-10}{a}=\frac{4}{a}$
$\Rightarrow \frac{t_2}{t_1}=\frac{4/a}{8/a}=\frac{1}{2}$