## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 5 - Problems - Page 187: 49

#### Answer

(a) The speed of the car is $17.7~m/s$ (b) The radial acceleration component of the car is $6.27~m/s^2$ (c) The magnitude of the total acceleration is $6.58~m/s^2$

#### Work Step by Step

(a) We can find the distance of $\frac{1}{4}$ of the circumference: $d = \frac{1}{4}\times 2\pi~r = \frac{1}{4}(2\pi)(50.0~m) = 78.54~m$ We can find the speed of the car after covering this distance: $v_f^2 = v_0^2+2ad$ $v_f = \sqrt{v_0^2+2ad}$ $v_f = \sqrt{0+(2)(2.00~m/s^2)(78.54~m)}$ $v_f = 17.7~m/s$ The speed of the car is $17.7~m/s$ (b) We can find the radial acceleration component of the car: $a_c = \frac{v^2}{r}$ $a_c = \frac{(17.7~m/s)^2}{50.0~m}$ $a_c = 6.27~m/s^2$ The radial acceleration component of the car is $6.27~m/s^2$ (c) We can find the magnitude of the total acceleration: $a = \sqrt{(2.00~m/s^2)^2+(6.27~m/s^2)^2} = 6.58~m/s^2$ The magnitude of the total acceleration is $6.58~m/s^2$

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