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Question

A vehicle can accelerate or decelerate at a maximum value of 1 ms−2 and can attain a maximum speed of 20 ms−1. If it starts from rest, what is the shortest time in which it can travel 1000 m, in a straight path, if it has to come to rest at the end of 1000 m?

A
50 s
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B
70 s
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C
30 s
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D
20 s
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Solution

The correct option is B 70 s
From the question we know that the vehicle starts from rest, attains a maximum speed of 20 m/s, and then comes to rest at the end.
This means the vehicle must have first accelerated to the maximum velocity and then decelerated to rest.

Let t1 and t2 be the time for acceleration (attaining maximum velocity) and deceleration (come to the rest) respectively.
Then, t1=Maximum velocity-Initial velocityacceleration
=2001=20 s
Then, distance travelled in time t1,
S1=ut1+12at21=0+12×1×202=200 m

Similarly, t2=0201=20 s
and S2=ut2+12at22=20×2012×1×202=200 m

Distance travelled at uniform velocity is
S=[1000(200+200)]
=600 m at 20 ms1
and time of travel for constant velocity is
t3=Sv=60020=30 s

Hence, total time for travel
=t1+t2+t3=70s.

It can be shown on vt graph as


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