AP Calculus multiple choice question The position of a particle along a line is given by $z(t) = 2t^3 -24t^2 + 90t +7$ for $ t \geq 0$. For what values of $t$ is the speed of the particle increasing?
a) $3 < t < 4$ only
b) $t > 4$ only
c) $t > 5$ only
d) $0 < t < 3$ and $t > 5$
e) $3 < t < 4$ and $t > 5$
For this problem I said d) based on my calculations:
$z'(t) = 6t^2 - 48t + 90 = 6(t-5)(t-3)$. I then did a chart and found out that $z'(0) > 0$ ,$z'(4) < 0$ and $z'(6) > 0$ leading me to conclude d) however the textbook says e). I thought that I did everything right, what is wrong with my conclusion? 
 A: First of all, realize that the speed is $|z^{'}(t)|$ and not $z^{'}(t)$. This is your mistake.
Now, graph $|z^{'}(t)|$ and you will see that your textbook is spot on.
A: The speed is increasing $\iff$ the acceleration and velocity have the same sign. The second derivative (acceleration) is given by $z''(t) = 12t - 48$, so $z''(t) > 0$ for $t > 4$ and $z''(t) < 0$ for $t < 4$.   Now it's easy to see why (e) is the answer.  
A: On a one-dimensional axis representing time, mark the signs of the particle's velocity and acceleration in each of the intervals determined by the zeros of the velocity and acceleration functions. The SPEED (not velocity) is increasing in any interval where the velocity and acceleration have the same sign. That is, the particle is accelerating while moving in what we define to be the positive direction or decelerating in what we define to be the negative direction.
A: A particle is speeding up when velocity and acceleration have the same signs and likewise is slowing down when velocity and acceleration have opposite signs. So you must take two derivatives. Technically you need to test the intervals created by all of the critical values in both derivatives. However since this is a MC question, just test those intervals into the derivatives. Use the factored form of each derivative for easy checking of positives and negative intervals.
