A particle is initially at rest and begins to accelerate to the right as 19.1t-t^2 m/s. How far does the particle move in 8.8 seconds?

Question

Var Acc

Answer 1

Probably the easiest way is to find the average velocity = (1/t) integral from 0 to t of a(t')dt' + v₀

In this case, v avg (t) = (1/t)(19.1t²/2 - t³/3) (0 limit is 0 when you plug it in and v₀ = 0) = 9.55t - t²/3

d = v_avg(t) t = v_avg(8.8s)(8.8s)

Please consider a tutor. Take care.

The usual way is to use v-v(0) = integral from 0 to t of adt and then d = integral from 0 to t of vdt.