Question

A thief in a stolen car passes through a police check post at his top speed of $90km{h}^{-1}.$ A motorcycle cop, reacting after $2s$, accelerates from rest at $5m{s}^{-2}.$ His top speed being $108km{h}^{-1}.$ Find the maximum separation between policemen and thief.

• A. $-112.5m$
• B. $115m$
• C. $116.5m$
• D. None of these

Answer 1

The correct option is A $-112.5m$

Using relative motion for solving the problem:
Assuming the thief is at rest and cop is moving at speed of $\Rightarrow v_{cop/thief}=v_{cop}-v_{thief}=0-90\times \frac{5}{18}=-25m/s$.
The distance travelled during first 5 s
$S_1=-25\ast 2=-50$
with relative acceleration of
$\Rightarrow a_{cop/thief}=a_{cop}-a_{thief}=5-0=5m/s^2$ in the forward direction.
For relative separation to be maximum the velocity of cop will be zero in opposite direction, then velocity will start increasing due to acceleration in positive direction, and the separation will keep on decreasing:
$v=u+at\Rightarrow 0=-25+5t$
$t=5s$
The distance travelled by cop during 5 s
$S=ut+\frac{1}{2}a{t}^2$
$S=-25\times 5+\frac{1}{2}\times 5^2$
$S=-62.5m$
maximum separation $=-50-62.5=-112.5$