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.

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/sPhysics

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