Question

Two object, $P$ and $Q$, travelling in the same direction starts from rest. While the object $P$ starts at time $t = 0$ and object $Q$ starts later at $t = 30$ min. The object $P$ has an acceleration of $40\,km/h^2$. To catch $P$ at a distance of $20\, km$, the acceleration of $Q$ should be

• A. $40 \, km/h^2$
• B. $80 \, km/h^2$
• C. $100 \, km/h^2$
• D. $120 \, km/h^2$
• E. $160 \, km/h^2$

Answer 1

Answer: (E)

According to question,
Object $P$ starts at $t=0$ and
acceleration of $P=40\, km / h ^{2}$
The time taken by object $P$ to cover $20\, km$ will be,
$s=u+\frac{1}{2} a t^{2}$
Where, $u=0, s=\frac{1}{2} a t^{2}$
$t=\sqrt{\frac{2 s}{a}}$
Here, $a=40\, km / h ^{2}$
and $s=20 \,kn$
$ t =\sqrt{\frac{2 \times 20}{40}} $
$ =1 $ hour
Time taken by $Q$ will be $\frac{1}{2}$ hour, because he
started after $30$ min.
The acceleration in $Q$,
$ a =\frac{2 s}{t^{2}}=\frac{2 \times 20}{\left(\frac{1}{2}\right)^{2}} $
$=2 \times 4 \times 20 $
$ =160 \,km / h ^{2}$