Question

Study the given paragraph(s) and answer the following questions.

A thief is running on a motorcycle at a constant speed of $25m{s}^{-1}$. A police jeep starts chasing from a point $1.25\mathrm{km}$ behind him with a uniform acceleration of $2{\mathrm{ms}}^{-2}$.

What should be the minimum acceleration of the thief to escape from the police.

• A. $1{\mathrm{ms}}^{-2}$
• B. $2.25{\mathrm{ms}}^{-2}$
• C. $1.5{\mathrm{ms}}^{-2}$
• D. $1.75{\mathrm{ms}}^{-2}$

Answer 1

The correct option is B

$2.25{\mathrm{ms}}^{-2}$

Step 1: Given

speed of thief $u_t=25{\mathrm{ms}}^{-1}$,

Distance $s=1.25\mathrm{km}=$ $1250\mathrm{m}$,

uniform acceleration of police $a_p=2{\mathrm{ms}}^{-2}$

Step 2 Solution

Let $a_t$ be the acceleration of the thief.
Relative acceleration between thief and police is given by,
$a_{pt}=a_p-a_t=2-a\dots \left(i\right)$
Using the equation of motion,
$v_{pt}^2=u_{pt}^2+2a_{pt}s$, here $v_{pt}$ is the final velocity of police with respect to
the thief,$u_{pt}$ is the initial velocity of police with respect to the thief
The thief will escape from the police. If $v_{pt}\le 0$.
$u_{pt}^2+2a_{pt}s\le 0$

Step 3: Calculation
From equation (i),
$(25{)}^2+2(2-a)\times 1250\le 0$
$⟹625+(4-2a)\times 1250\le 0$
$⟹625+5000-2500a\le 0$
$⟹a\ge \frac{5625}{2500}$
$⟹a\ge 2.25{\mathrm{ms}}^{-2}$
The minimum acceleration of the thief to escape from the police is$2.25{\mathrm{ms}}^{-2}$.

Hence option B is correct.