Question

The acceleration a of a particle starting from rest varies with time is given by a (2t-6) find time in seconds at which velocity of particle in negative direction is maximum

Answer 1

Given : The acceleration a of a particle starting from rest varies with time is given by a (2t-6)

To Find :  time in seconds at which velocity of particle in negative direction is maximum

Solution:

a = 2t - 6

a = dv/dt

=> dv/dt = 2t - 6

=> dv = (2t - 6) dt

integrating both sides

=> v  =  t² - 6t  + C

particle starting from rest  => at t = 0 , v = 0

=> 0  =  0 - 0 + C

=> C = 0

v  =  t² - 6t

=> v =  t² - 6t

t² - 6t + 9 - 9

= ( t - 3)² - 9

Hence velocity of particle in negative direction is maximum

at t = 3

time in seconds at which velocity of particle in negative direction is maximum is 3 sec

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