# Thread: Rate of change help

1. ## Rate of change help

The height (in m) that an object has fallen from a height of 180 m is given by the position function s(t) = -5tsquared + 180, where t > or equal to 0 and t is in seconds.

a. Find the average velocity during each of the first 2 seconds.

b. Find the velocity (instantaneous) of the object at t = 4

c. At what velocity will the object hit the ground?

I tried these, and got different answers then the correct ones. Help me out?

Velocity at a moment? That's what a derivative is for.

Average Volocity over time? That's what integrals are for.

When does it hit the ground? That's what the position function is for.

Let's see how you managed what you managed.

3. Originally Posted by Hockey_Guy14
The height (in m) that an object has fallen from a height of 180 m is given by the position function s(t) = -5tsquared + 180, where t > or equal to 0 and t is in seconds.

a. Find the average velocity during each of the first 2 seconds.

b. Find the velocity (instantaneous) of the object at t = 4

c. At what velocity will the object hit the ground?

I tried these, and got different answers then the correct ones. Help me out?
a) The average velocity between times $\displaystyle t_1$ and $\displaystyle t_2$ is found by the formula $\displaystyle \frac{s_2 - s_1}{t_2 - t_1}$.

b) The instantaneous velocity can be found by taking the first time derivative of the position function s(t).

c) When does the object hit the ground? When s(t) = 0, of course. So solve for t and plug that into the velocity equation you got for b.

-Dan