The position function of an object moving along a straight line is given bys = f

Question

The position function of an object moving along a straight line is given bys = f(t).The average velocity of the object over the time interval [a, b] is the average rate of change of f over [a, b]; its (instantaneous) velocity at t = a is the rate of change of f at a.

A ball is thrown straight up with an initial velocity of 96 ft/sec, so that its height (in feet) after t sec is given by s = f(t) = 96t ? 16t2.

(a) What is the average velocity of the ball over the following time intervals [4,5] [4,4.5] [4,4.1] ?

Explanation / Answer

height is given by s = 96t - 16 t^2

then f(t) = 96t - 16 t^2
instantaneous velocity = df/dt = 96 - 32t
average velocity( [4,5]) = total displacement / total time = 96 - 16(9) = -48 / 1 = -48
   average velocity( [4,4.5]) = total displacement / total time = (96(0.5) - 16(4.25)) / 0.5 = -20 / 0.5 = -40
average velocity( [4,4.1]) = total displacement / total time = (96(0.1) - 16(0.81)) / 0.1 = -3.36 /0. 1 = -33.6

total displacement [t1,t2] = f(t2) - f(t1) in [4,5] = 96(5) - 16*(5^2) - [ 96(4) -16(4^2) ] = 96 - 16(9) = -48

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