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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)
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