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
Answers
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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