displacement question?

Question

The displacement 's' metres at time 't' seconds is given by:

s = 5cos3t + t^2 + 10 , for t => 0

a) Write doen the minimum value of 's'.

b) Find the acceleration, 'a', at time 't'.

c) Find the value of 't' when the maximum value of 'a' first occurs.

Answer 1

ds/dt = -15sin 3t +2t = 0
With a calculator you get t=0 and s=1.00251
the minimum is s=6.05 m
a= -45cos3t +2
da/dt = 145 sin 3t = 0 so 3t = kpi and t = k*pi/3
for t= 0 a=-43 m/s^2
and k= 1 t=pi/3 (first maximum) and a = 47m/s^2

Answer 2

s'= -15sin3t+2t =0 at t=0

smin=6.1
s'' = a = -45cos3t + 2
s'''= a' = 135sin3t=0 at t = 0 at t=60,120 and is first max at t=60 or pi/3. At 60 because the slope is positive before t = 60 and negative after