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Question

A car is travelling on a straight road. The maximum velocity the car can attain is 24 m/s. The maximum acceleration and deceleration it can attain are 1 ms2 and 4 ms2 respectively. The shortest time( in s) the car takes from rest to rest in a distance of 480m is:

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Solution

Using
t=1v+v2(1α+1β)
=48024+242(11+14)
=20+12×54=20+15=35 s

OR

Distance covered to attain maximum possible speed with maximum acceleration is S1=V22a=2422=288 m
Time taken t1=va=24 s
Distance covered when car reaches from maximum possible speed to rest is S2=V22a=2422(4)=72 m
Time taken t2=va=24/4=6 s
S1+S2=360m

Given distance is 480 m, hence remaining 120 m is covered at maximum speed,
For which time taken is t3=12024=5 s

Total time taken is t1+t2+t3=35 s

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