A weight oscillates in a vertical motion according to the position function y(t) = –5 cos t. Assuming t ≥ 0

6) A weight oscillates in a vertical motion according to the position function y(t) = –5 cos t. Assuming t ≥ 0, when will the acceleration of the weight be zero for the first time?

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.