A driver applies brakes to the vehicle on seeing traffic signal 400 m ahead .At time of applying the brakes vehicle was moving with 15m/s and then starts retarding with 0.3 m/s^2 .The distance of vehicle after 1 min from traffic light A : 25 mB : 375mC : 360mD ; 40m

The equations of motion are:

Using relation V² - u² = 2*a*S

      and Using relation V - U = a*t     V:final vel.  U: initial vel.  a:accln  S:distance travelled

The distance travelled before car comes to rest is V=0, U = 15,a=-0.3

S=225/0.6 =375mts

The car comes to rest even before reaching traffic signal.

It comes to rest at 25m ahead traffic signal.

The time taken by car to come to rest is : V/a = 15/0.3 = 50 sec

In 50 seconds car comes to rest

so after 1 min,also car stays 25m ahead of traffic post

option A

Distance travelled by vehicle in 1 min.(=60 sec.) after applying brakes=15*60-(0.3)*60² = -180mtrs.(s=ut+1/2at²)                                    

 (-ve) value above indicates that vehicle must've stopped due to brake action in less than 1 min.

So using,

                v²=u²+2as;s=(15)² /(2*0.3)=375m

So answer would be A:25m(400-375)