Question

Two objects $P$ and $Q$ , travelling in the same direction start 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. $160 \, km / h^{2}$

Answer 1

Answer: (D)

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,
$s=ut+\frac{1}{2}at^{2}$
Where, $u=0, \, s=\frac{1}{2}at^{2}$
$t=\sqrt{\frac{2 s}{a}}$
Here, $a=40 \, km/h^{2}$
and $s=20 \, km$
$t=\sqrt{\frac{2 \, \times \, 20}{40}}$
$=1 \, hour$
Time taken by $Q$ will be $\frac{1}{2} \, hour$ , because it started after $30 \, min$ .
The acceleration in $Q$ ,
$a=\frac{2 s}{t^{2}}=\frac{2 \, \times \, 20}{\left(\frac{1}{2}\right)^{2}}$
$=2\times 4\times 20$
$=160 \, k m / h^{2}$