Calculus Volume 1

496

Chapter 4 | Applications of Derivatives

Integrating, we have

t 2 +88 t + C .

s ( t ) = − 15 2

Since s (0) =0, the constant is C =0. Therefore, the position function is s ( t ) = − 15 2 t 2 +88 t .

sec, the position is s ⎛

⎞ ⎠ ≈258.133 ft.

⎝ 88 15

After t = 88 15

4.53 Suppose the car is traveling at the rate of 44 ft/sec. How long does it take for the car to stop? How far will the car travel?

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