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 jumpers velocity is approximated by v(t) 4t 2 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 5 0 v(t) dt represent in terms of the jump

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,

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 approximated 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 ∫⁵₀ v(t) dt represent in terms of the jump?

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a Solving vt 4t 2 16t 0 gives t 0 and t 4 At t 0 the jump has just started at t 4 the jumper is mo... View full answer

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