Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 2 - Describing Motion: Kinematics in One Dimension - Problems - Page 44: 32

Answer

The stopping distance is 50.7 m. She will not be able to stop in time. She will be at a distance of 30.7 m past the beginning of the intersection.

Work Step by Step

For the first 0.350 s, the acceleration is 0 because this is the driver's reaction time. We can calculate the distance traveled in the first 0.350 s: $x = v_0t = (18.0 ~m/s)(0.350 ~s) = 6.30 ~m$ We can use $x_0 = 6.30 ~m$ in the next part of the question. $x = x_0 + \frac{v^2-v_0^2}{2a} = 6.30 ~m + \frac{0-(18.0 ~m/s)^2}{(2)(-3.65 ~m/s^2)} = 50.7 ~m$ The stopping distance is 50.7 m. Since the driver is 20.0 m from the intersection, she will not be able to stop in time. She will be at a distance of 30.7 m past the beginning of the intersection.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.