## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 357: 75

#### Answer

A constant acceleration of $5.87~ft/s^2$ is required.

#### Work Step by Step

We can express $30~mi/h$ in units of $ft/s$: $(30~mi/h)\times \frac{5280~ft}{1~mi}\times \frac{1~h}{3600~s} = 44~ft/s$ We can express $50~mi/h$ in units of $ft/s$: $(50~mi/h)\times \frac{5280~ft}{1~mi}\times \frac{1~h}{3600~s} = 73.33~ft/s$ $a(t) = \frac{dv}{dt} = a$ $v(t) = v_0+a~t$ When $t = 0$, then $v = 44~ft/s$ Thus $v_0 = 44~ft/s$ $v(t) = 44+a~t$ We can find $a$: $v(5) = 44+a~(5) = 73.33$ $5a = 29.33$ $a = 5.87~ft/s^2$ A constant acceleration of $5.87~ft/s^2$ is required.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.