To get the speed of the particle at any time, simply take the magnitude of the velocity vector. In 1 - D motion this corresponds to taking the absolute value of the velocity.
s(t) = 2t^3 - 24t^2 + 90t + 7
v(t) = 6t^2 - 48t + 90 <-- This is the velocity.
We wish to know for what times |v(t)| is increasing. For this we need the derivative of |v(t)| = |6t^2 - 48t + 90|
By the chain rule:
d|v(t)|/dt = sign(6t^2 - 48t + 90)*(12t - 48)
When is this positive?
We need to know when sign(6t^2 - 48t + 90) > 0
The zeros of 6t^2 - 48t + 90 are:
6t^2 - 48t + 90 = 0
t^2 - 8t + 15 = 0
(t - 3)(t - 5) = 0
t = 3 s, 5 s
So for t in the domain
[0, 3), 6t^2 - 48t + 90 > 0
(3, 5), 6t^2 - 48t + 90 < 0
(5, infinity), 6t^2 - 48t + 90 > 0
Thus sign(6t^2 - 48t + 90) > 0 for t belonging to [0, 3) and (5, infinity) and sign(6t^2 - 48t + 90) < 0 for t belonging to (3, 5).
Now, 12t - 48 is positive for t >4.
Putting it all together we get that:
d|v(t)|/dt > 0 for t belonging to (3,4) since both sign() and 12t - 48 are both negative. For the time period t > 5 s, both sign() and 12t - 48 are positive, so the speed is also increasing for t > 5 s.
Thus the speed is increasing for the time period 3 s to 4 s and t > 5 s, which is answer E.
This can all be done without a calculator (I didn't use one) but it IS a long question to answer.