Verified answer

Remember the acceleration is defined to be the 2nd derivative of the position function:

y(t) = -5cost

y '(t) = 5sint = velocity

y ''(t) = 5cost = acceleration

Set the acceleration equal to 0 and solve for t:

0 = 5cost

cost = 0

t = kπ, where k is 0, 1, 2, 3, ...

So the first time that the acceleration will be 0 is when k =0. Which means:

t = (0)π = 0

So at t = 0 is the first time that the acceleration is equal to 0.

y(t)=-5cos t

y'(t)= 5 sin t

y''(t) = acceleration = 5 cos t

5 cos t = 0

cos t = 0

t = pi/2

Acceleration is 0 when t=1.57 seconds for the first time.