Extreme need of online calculus help that is due tonight.....please help!?

Question

The famous but unlucky cartoon character Wile E. Coyote falls from a tall cliff while chasing the Road Runner. The Coyote's height above the ground, in feet, can be described by the position function s(t)= -16t^2+350, where time t is in seconds.

-Find the Coyote's velocity function v(t).

-What is the Coyote's velocity after he has been falling for two seconds?

-What is the Coyote's average velocity for the time interval [0,2]?

-As the Coyote begins to fall, the Road Runner races to the bottom of the cliff, taking 5 seconds to get there. How does the Road Runner get there in time to see Coyote hit the ground? HOW DO YOU KNOW?

Can someone please tell me the steps to get the answers?

Answer 1

Velocity is the derivative of position (and acceleration is the derivative of velocity). If our position function is given as above, then we can take that derivative to find Mr. Coyote's velocity at any time t:

s(t) = -16t^2 + 350
v(t) = -32t

Now that we have this, we can plug in t = 2 to find his velocity:

v(2) = -32(2)
v(2) = -64 ft/s

The negative indicates that the poor guy is falling...as usual.

Finding the average velocity is easy:

v(avg) = [v(2) - v(0)] / [2 - 0]
v(avg) = [-64 - 0] / [2]
v(avg) = -32 ft/s

Finally, we see that our speedy friend is going to beat the coyote to the bottom of the cliff. Just for kicks, we can figure out how long the coyote takes to get there by setting s(t) = 0:

s(t) = -16t^2 + 350
0 = -16t^2 + 350
16t^2 = 350
t^2 = 21.875
t = 4.677 s

So if the Roadrunner takes 5 seconds, there's no way that he can get there in time to see his foe go splat.