Question

The displacement x( in m) of a body varies with time (in s) as x = (t^2) - 16t +2 . At what time does the body come to rest?

Answer 1

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here x= 16t− 2t2 in such type of questions here take care of the point where particle take u turn v = dx/dt = 16−4t a=dv/dt = −4 here initialy velocity is positive and acceleration is negative so particle take U−turn when its velocity become zero now 16−4t=0 gives t=4seconds . so particle will take u turn at 4 seconds position of particle at t=2 seconds =16*2−2*2*2=24 postion at t=4seconds = 16*4−2*4*4=32again position at t=6sec =16*6−2*6*6=24 now particle again come at x=24 position after takig u turn so distance travelled in 6 seconds= 24+8+8 = 40 now 3S2S1= 3*40/24 =5 answer

Answer 2

Answer: 8 s

The body will come to rest at t=8s

Explanation:

x= t^2-16t+2

We have to find the time at which v=0...

v=dx/dt

dx/dt= d(t^2)/dt - d(16t)/dt + d(2)/dt

》v= 2t-16

When v=0,

2t=16

t=8

Hence the required time is 8 s.