Displacement, velocity and acceleration problems- questions d, e, and f

(d) The body is stationary when it has no velocity (when it's not moving).

Velocity is the measure of the rate of change of displacement,
v = s'

s = t^3 + 2t^2 - 4t - 12
Differentiating, applying our power rule to each term,
s' = 3t^2 + 4t - 4

This is our velocity function.
v = 3t^2 + 4t - 4
We want to know when the velocity is zero,
0 = 3t^2 + 4t - 4

Solve for t. You should end up with two values.
A negative t doesn't make sense within the context of the problem.
So disregard that value.

(e) Acceleration is the rate of change of velocity.
So again we take a derivative,

s'' = 6t + 4
This is our acceleration function.
a = 6t + 4
Evaluate this function at t=3.

(f) Remember: The body is stationary when it's velocity = 0. Use your t value that you found in part (d) and plug it into the acceleration function.

The question is kind of confusing. if it's asking how many variations of answers can you give. then do 4^5 for your answer.this would be something like the first question could be a and the 2nd c or first b next c etc. if you can, ask the teacher for a better confirmation of what it's asking. if i'm assuming correctly what it's asking, it's 625 different variations.