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Question

The acceleration of a particle starting from rest vary with respect to time is given by a = (2t - 6), where t is in seconds. Find the time at which velocity of particle in negative direction is maximum.

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Solution

a=dv/dt

a=2t+6

dv=a*dt

dv=(2t-6)dt

integrating both sides we get

V=2t^2/2-6t =t^2-6t


velocity =0 when t^2-6t=0

=>t(t-6)=

=>either at t=0 or t=6

So velocity is negative between t=0 and t=6


=>when a=0

=>velocity in negative direction will be maximum

=>2t-6=0

=>t=3

Velocity in negative direction is maximum when t=3s

and maximum negative velocity=t^2-6t=9-18= -9m/s


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