The distance s (in feet) covered by a motorcycle traveling in a straight line an

Question

The distance s (in feet) covered by a motorcycle traveling in a straight line and starting from rest in t sec is given by the function
s(t) = 0.1t^3 + 4t^2 + 25t
Calculate the motorcycle's average velocity over the time interval [2, 2 + h] for h = 1, 0.1, 0.01, 0.001, 0.0001, and 0.00001. (Round your answers to four decimal places.)
h = 1

h = 0.1

h = 0.01

h = 0.001

h = 0.0001

h = 0.00001

Use your results to guess at the motorcycle's instantaneous velocity at t = 2. (Round your answer to one decimal place.)

Explanation / Answer

Your hint might mislead - it should be: average velocity V over the interval 2 to (2 + h) seconds V = [f(2 + h) - f(2)]/(2 + h) - 2] = [f(2 + h) - f(2)]/h The calculated values for different lengths of the time interval after 2.0 s, with s(2) = 57.20 ft, were . h . . . 0.1 . . . . . . 0.01 . . . . . .0.001 . . . . . . .0.0001 . . . . . . . .0.00001 s s(t) . . 60.3939 . . .57.51814 . . 57.2318014 . . 57.203180014 . . .57.20031800014 ft . V . . .31.939 . . . 31.814 . . . . 31.8014 . . . . .31.80014 . . . . . . 31.800014 ft/s The value for the instantaneous velocity at t = 2.0 s obtained by differentiating s(t) is V = ds/dt = - 0.3.t² + 4.t + 25 = 31.8 ft/s The differences from this shown in the bottom row of the table originate in the change in velocity over the small but finite time interval. We may calculate this from the first equation in this answer to be V(h) = 31.8 + 1.4.h - 0.1.h² The last term has been rounded out of all values except the first.

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