The stopping distance for an automobile (after applying the brakes) is approxima

Question

The stopping distance for an automobile (after applying the brakes) is approximate F(s) = 1.1s + 0.054s^ ft. where 8 is the speed in mph. Use the Linear Approximation to estimate the change in stopping distance per additional mph when s = 40 and when s = 65. (Use decimal notation. Give your answer to three decimal places.) The change in stopping distance per additional mph for s = 40 mph is approximately ft. help The change in stopping distance per additional mph for s = 65 mph is approximately Note: you can earn partial credit on this problem.

Explanation / Answer

Answer:

The stopping distance of an automobile(after applying brakes ) is approximately f(s)=1.1s+0.054s^2

where s is the speed in mph

Then f'(x) = 1.1+0.108s

By Linear approximation we have
f(x+h) f(x) + f '(x) h
take h=1 to know for each 1 additional mph
when s=40 we have
f(40+1) f(40) + f '(40)(1)
f(40+1) [1.1(40) +.054(40)^2] + [1.1 + .108(40)](1)
f(40+1) [44 + .054(1600)] + [1.1+4.32]
f(40+1) 130.4 + 5.42
f(40+1) 135.82 ft

(b)

Now when s=65, we have

f(65+1) f(65) + f '(65)(1)
f(65+1) [1.1(65) +.054(65)^2] + [1.1 + .108(65)](1)
f(65+1) [71.5 + .054(4225)] + [1.1+7.02]
f(65+1) 299.65 + 8.12
f(65+1) 307.77 ft

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