Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 2 - Kinematics in One Dimension - Exercises and Problems: 10

Answer

The acceleration during the speeding up phase is $16 ~m/s^2$ The acceleration during the slowing down phase is $-5.33 ~m/s^2$

Work Step by Step

The acceleration is equal to the slope of the velocity versus time graph. We can find the slope from point A to point B when the blood is speeding up. We can use point A (0.1 s, 0 m/s) and point B (0.15 s, 0.8 m/s) to calculate the slope. $a_y=\frac{\Delta v_y}{\Delta t}$ $a_y = \frac{0.8 ~m/s - 0}{0.15 ~s-0.1~s}$ $a_y = \frac{0.8 ~m/s}{0.05 ~s}$ $a_y = 16 ~m/s^2$ The acceleration during the speeding up phase is $16 ~m/s^2$ We can find the slope from point B to point C when the blood is slowing down. We can use point B (0.15 s, 0.8 m/s) and point C (0.3 s, 0 m/s) to calculate the slope. $a_y=\frac{\Delta v}{\Delta t}$ $a_y = \frac{0-0.8 ~m/s}{0.3 ~s-0.15~s}$ $a_y = \frac{-0.8 ~m/s}{0.15 ~s}$ $a_y = -5.33 ~m/s^2$ The acceleration during the slowing down phase is $-5.33 ~m/s^2$
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