As a certain object falls, its position s (in meters) above ground after t secon

Question

As a certain object falls, its position s (in meters) above ground after t seconds is given by s(t) = 30 - 5t^2. What is the average velocity of the object on the interval from t = 1 to the time 0.5 seconds later? m/s What is the average velocity of the object on the interval from t = 1 to the time 0.1 seconds later? m/s Write a polynomial in terms of h for the average velocity from t = 1 to the time h seconds later (h notequalto 0). (Simplify your answer completely.) What does this average tend toward for h closer and closer to 0 (smaller and smaller time interval)? m/s What would you guess to be the exact velocity of the object after 1 second? m/s

Explanation / Answer

s(t) = 30 - 5t^2

time interval is t E [1 , 1+h]

average velocity v(h) = [v(1+h) - v(1)]/(1+h - 1)

                                    = {[30-5(1+h)^2] -[30 - 5(1)^2 ]}/h

                                   v(h) = (-5h^2 -10h)/h = -5h - 10

d> as h --> 0 v(h) tends towards -10

lim h -->0 (-5h - 10) = -5(0) - 10 =-10 m/s

answer is -10 m/s

e> s(t ) = 30 -5t^2

v(t) =ds/dt = -10t

v(t = 1) = -10(1) = -10 m/s

=> v(t=1) = -10 m/s

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