The "position" of a particle is a code name for the "displacement" of a particle, which is the vector version of distance from the origin. So when you take the derivative of the position function you get the velocity of the particle.

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.

So:

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.

-Dan