Question: A bungee jumper leaps off the starting platform at time t = 0 and rebounds once during the first 5 seconds. With velocity measured downward, for t in seconds and 0 ≤ t ≤ 5, the jumper's velocity is approximated5 by v(t) = -4t2 + 16t meters/sec.
(a) How many meters does the jumper travel during the first five seconds?
(b) Where is the jumper relative to the starting position at the end of the five seconds?
(c) What does 0∫5 v(t) dt represent in terms of the jump?