Thread: I need help with this word problem in calculus. pretty please

The equation for free fall at the surface of the moon is s=31.8t^2in with t in sec. Assume a rock is dropped from the top of a 12,000 in cliff. Find the speed of the rock at t=4sec. Iḿ not sure if Iḿ over complicating things, but I do not know if this is plug or chug or if it is more complicated than that, because we have been working on limits and this has nothing to do with this.

What do you get if you differentiate s with respect to t? What does this represent?

Originally Posted by wbarner

The equation for free fall at the surface of the moon is s=31.8t^2in with t in sec. Assume a rock is dropped from the top of a 12,000 in cliff. Find the speed of the rock at t=4sec. Iḿ not sure if Iḿ over complicating things, but I do not know if this is plug or chug or if it is more complicated than that, because we have been working on limits and this has nothing to do with this.

The equation $s=31.8t^2\text{ in}$ means that at time $t$, the rock is at position $s$ inches. Let's determine how far the rock travels in 4 seconds (if it has hit the bottom of the cliff, that will certainly affect its speed).

$$s = 31.8(4)^2 = 508.8\text{ in} < 12,000\text{ in}$$

So, the rock is still falling after four seconds. Next, we have the point $(4,508.8)$. If we consider another point, $(t,s) = (t,31.8t^2)$, the average speed of the rock between those two points is the slope of the line between those two points. It is the change in distance over the change in time. This is written as $\dfrac{\Delta s}{\Delta t}$. The closer we get the second point $(t,s)$ to the point $(4,508.8)$, the closer this average becomes to the instantaneous rate of change at $t=4$. In other words, we are looking for the following limit:

$$\lim_{t \to 4} \dfrac{s-508.8}{t-4} = \lim_{t \to 4} \dfrac{31.8t^2-508.8}{t-4}$$

Can you solve it from here?