find the body's speed & acceleration of S= -t^3 + 3t^2 - 3t @ the endpoints of the interval 0<_ t <_ 3

Question

find the body's speed & acceleration of S= -t^3 + 3t^2 - 3t @ the endpoints of the interval 0<_ t <_ 3

Answer 1

speed= I dS/dtI = I -3t^2+6t-3I
at t=0 = 3 and at t=3 = 12
acceleration = d(dS/dt)/dt = -6t +6
at t=0 = 6 and at t=3= -12

Answer 2

Speed is the rate of change of distance w.r.t time, so:

s = -t^3 + 3t^2 -3t, so

ds/dt = v = -3t^2 + 6t -3, and

acceleration is the rate of change of speed w.r.t time, so:

v = -3t^2 + 6t -3, so

dv/ds = a = -6t + 6.

Hope this helps, Twiggy.