If we ignore air resistance, a falling body will fall 16t2 feet in t seconds. What is the average velocity between t

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1. Answer:

   262.4 m/s

   Explanation:

   The complete question is

   If we ignore air resistance, a falling body will fall 16t^2 feet in t seconds. What is the average velocity between t=8 and t=8.4? Round your answer to two decimal places if necessary.

   The distance fallen s = 16t^2

   The velocity v = [tex]\frac{ds}{dt}[/tex] = 32t

   If we substitute the values of t into the velocity v, we’ll have

   at t = 8 s,     V1 = 32 x 8 = 256 m/s

   at t = 8.4 s,  V2 = 32 x 8.4 = 268.8 m/s

   Average velocity = (V2 – V1)/2 = (268.8 + 256)/2 = 262.4 m/s

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