Question #86f49

Oct 19, 2016

$15 s$
(not there in the given choices).

Explanation:

Let the motor cycle catch up with the car in $t$ second.
Distance moved by car during this period $s = {v}_{c a r} \times t$
$s = 24 \times t = 24 t \text{ m}$

Kinematic equation for motor cycle is
$s = u t + \frac{1}{2} a {t}^{2}$
The distance moved by car must be equal to the distance moved by motor cycle. Inserting calculated and given values we get
$24 t = 0 \times t + \frac{1}{2} \times 3.2 {t}^{2}$
$\implies 24 t = 1.6 {t}^{2}$
Gives the values $t = 0 \mathmr{and}$
$24 = 1.6 t$
$\implies t = \frac{24}{1.6} = 15 s$
Ignoring $t = 0$, which represents initial point when car crossed the stationary motor cycle.