Help with this please..?

Question

The velocity of a particle t s after it starts from rest is v m s ^-1, where v = 1.25t - 0.05t^2. find the displacement of the particle from its starting point at the instant when its acceleration is 0.05m s^-2.

Answer 1

dv/dt=1.25-.1t
This equals .05 when .05=1.25-.1t or when -1.2=-.1t
or when t=12 seconds. Between 0 and 12 seconds, you must integrate the velocity function between t=0 and t=12 seconds.The integral of the velocity function is .75t^2-(.05/3)t^3. I will let you do the rest.