Question #86f49

1 Answer
Oct 19, 2016

Answer:

#15s#
(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_(car)xxt#
#s=24xxt=24t" m"#

Kinematic equation for motor cycle is
#s=ut+1/2at^2#
The distance moved by car must be equal to the distance moved by motor cycle. Inserting calculated and given values we get
#24t=0xxt+1/2xx3.2t^2#
#=>24t=1.6t^2#
Gives the values #t=0 and#
#24=1.6t#
#=>t=24/1.6=15s#
Ignoring #t=0#, which represents initial point when car crossed the stationary motor cycle.