Essential University Physics: Volume 1 (4th Edition)

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

Chapter 3 - Exercises and Problems - Page 50: 18


Please see the work below.

Work Step by Step

As we know that $\Delta v=v_2-v_1$ $\Delta v=\sqrt{v_1^2+v_2^2-2v_1v_2cos(\theta)}$ We plug in the known values to obtain: $\Delta v=\sqrt{(15)^2+(19)^2-2(15)(19)cos(28)}$ $\Delta v=9.095\frac{Km}{s}$ $\Delta v=9095\frac{m}{s}$ Now we can find the acceleration as $a=\frac{\Delta v}{\Delta t}$ $a=\frac{9095}{10\times 60}=15\frac{m}{s^2}$
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