Problem of the Week #11

Problem 11: 1998-1999

The Furby you got during the holiday is beginning to annoy you. In fact, you no longer toh-loo (like) the little stinker. So, to play a little loo-loo (joke), you drop him from a height of 300 feet straight nah-bah (down). Its velocity after t seconds is -32t ft/sec. Perhaps now he'll be quiet and way-loh koh-koh (sleep again).

(a)    Determine the position function.
(b)    How fast is the Furby dropping after 4 seconds?
(c)    How far has the little guy dropped after 4 seconds?
(d)    How many seconds will it take for the famous Furby Splat?

Solution: (a) You know that the position function, s(t), is the integral (antiderivative) of velocity. Therefore, s(t)=-32t²/2 + C. We also know that initial (t = 0) position is 300. Therefore, 16(0)² + C = 300, and C=300. Finally, s(t) = -16t² + 300.

(b) v(t)=-32t, so v(4) = -32(4) = -128 ft/sec, or 128 ft/sec down.

(c) s(4) = -16(4)²+ 300 = -16(16) + 300 = -256+300 = 44 feet off the ground at t = 4, so it has dropped 300-44 = 256 feet.

(d) When the Furby hits the ground ("me no like"), its position will be zero feet off the ground, clearly. So, we must solve the equation -16t² + 300 = 0. Subtract 300 from both sides, divide by -16 and take the square root to get 4.330 seconds as Furby touchdown time. Shame on you, he just wanted tickled.