University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 2 - Motion Along a Straight Line - Problems - Exercises - Page 60: 2.27

Answer

(a) The acceleration is $3.1\times 10^6~m/s^2$, which is $3.2\times 10^5g$ (b) This acceleration would last 1.6 ms. (c) Because $3.2\times 10^5g$ is less than $4.5 \times 10^5 g$, we can not rule out the possibility that our ancestors were Martian microbes.

Work Step by Step

(a) We can calculate the acceleration. $a = \frac{v^2-v_0^2}{2x} = \frac{(5000~m/s)^2-0}{(2)(4.0~m)} = 3.1\times 10^6~m/s^2$ The acceleration is $3.1\times 10^6~m/s^2$ $a = \frac{3.1\times 10^6~m/s^2}{9.80~m/s^2} = 3.2\times 10^5g$ (b) $t = \frac{v-v_0}{a} = \frac{5000~m/s-0}{3.1\times 10^6~m/s^2} = 0.0016~s = 1.6~ms$ This acceleration would last 1.6 ms (c) Because $3.2\times 10^5g$ is less than $4.5 \times 10^5g$, we can not rule out the possibility that our ancestors were Martian microbes.
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