Question

Question #86f49

Answer 1

`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.