# The Derivative as a Rate of Change

• Oct 10th 2009, 09:29 PM
VitaX
The Derivative as a Rate of Change
A rock thrown vertically upward from the surface of the moon at a velocity of $24$ $\frac{m}{sec}$ reaches a height of $s=24t-0.8t^2$ meters in $t$ $sec$.
a. Find the rock's velocity and acceleration at time $t$ $\rightarrow v=24-1.6t$ $\frac{m}{s}$ $a=-1.6$ $\frac{m}{s^2}$
b. How long does it take the rock to reach its highest point? $\rightarrow 24-1.6t=0$ $t=15$ $sec$
c. How high does the rock go? $\rightarrow s_{max}=24(15)-0.8(15)^2$ $s_{max}=180$ $m$
d. How long does it take the rock to reach half its maximum height?
e. How long is the rock aloft? $\rightarrow 30$ $sec$

I don't know how to find how long it takes the rock to reach half its max height. How would I solve this?
• Oct 10th 2009, 09:58 PM
VonNemo19
Quote:

Originally Posted by VitaX
I don't know how to find how long it takes the rock to reach half its max height. How would I solve this?

Well, once ya find the max height (call it $s_{max}$) then throw $\frac{s_{max}}{2}$ into the left side of your displacement function and then solve for $t$.

(Wink)
• Oct 10th 2009, 10:09 PM
VitaX
Quote:

Originally Posted by VonNemo19
Well, once ya find the max height (call it $s_{max}$) then throw $\frac{s_{max}}{2}$ into the left side of your displacement function and then solve for $t$.

(Wink)

(Speechless) overlooked that easy bit. So when I put them in the quadratic formula, do I disregard one answer or take both as when its going up and then going down?
• Oct 10th 2009, 10:13 PM
VonNemo19
Quote:

Originally Posted by VitaX
(Speechless) overlooked that easy bit. So when I put them in the quadratic formula, do I disregard one answer or take both as when its going up and then going down?

Good question. I thought about that when I answered your post. I would tend to say that the question is concerned about the upward motion, but I wouldn't neglect the downward.

I would use both, but I would explain my answer to avoid confusion and cover my bases.