Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 2 - Exercises and Problems - Page 32: 70


$8.59 \ km/h$

Work Step by Step

We know the following equation: $x = v_0t + \frac{1}{2}at^2$ Thus, we find: $x_{20}+v_{20}t= v_{10}t + \frac{1}{2}at^2$ $0= v_{10}t + \frac{1}{2}at^2-x_{20}-v_{20}t$ Solving for t by applying the quadratic formula, we plug in the known values to find: $t = \frac{(23.61-16.67)\pm\sqrt{(23.61-16.67)^2-4(\frac{-1}{2})(4.2)(10)}}{-4.2\times 2 \times \frac{1}{2}}=1.085 \ seconds$ Thus, we find the velocity of the second car: $v = v_1 - 60\ km/h$ $v = (23.61-(4.2)(1.085))(3.6 \frac{km/h}{m/s}) - 60\ km/h=8.59 \ km/h$
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