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 - Page 68: 80

Answer

a) $8.13\;\rm m/s^2,\; 2.05\;m/s^2 ,\; 0.517\;m/s^2$ b) $x=17.148\left( e^{-0.6887t} +0.6887t-1\right)$ c) $9.9\;\rm s$

Work Step by Step

a) To find the acceleration we need to differentiate the $v_x$ relative to $t$. $$a_x=\dfrac{d}{dt}v_x=\dfrac{d}{dt}a(1-e^{-bt})$$ $$a_x =a\dfrac{d}{dt}(1-e^{-bt})=ab e^{-bt}$$ Plugging the constant magnitudes $$a_x=(11.81\cdot 0.6887) e^{-0.6887t}=8.1335\;e^{-0.6887t}$$ $$\boxed{a_x=8.1335\;e^{-0.6887t}}\tag 1$$ Thus, at $t=0\;\rm s$, $$a_x= 8.1335\;e^{-0.6887\times 0}=\color{red}{\bf 8.1335}\;\rm m/s^2$$ at $t=2\;\rm s$, $$a_x= 8.1335\;e^{-0.6887\times 2}=\color{red}{\bf 2.052}\;\rm m/s^2$$ at $t=4\;\rm s$, $$a_x= 8.1335\;e^{-0.6887\times 4}=\color{red}{\bf 0.5175}\;\rm m/s^2$$ b) To find the distance, we need to take the integral of $v_x$ relative to $t$. Thus, the left side of the given formula $$\int v_x dt=\int \dfrac{dv_x}{dt}dt=\int_0^x dx=x$$ And hence, the right side, $$ x=\int_0^t(a-ae^{-bt})dt=\int_0^t adt-\int_0^t ae^{-bt}$$ $$ x=at\bigg|_0^t-\left( \dfrac{ae^{-bt}}{-b}\right)\bigg|_0^t$$ $$ x=at+\left( \dfrac{ae^{-bt}}{ b}-\dfrac{a}{b}\right) $$ $$ x=at+\dfrac{a}{b}\left( e^{-bt} -1\right) $$ $$\boxed{ x= \dfrac{a}{b}\left( e^{-bt} +bt-1\right) }$$ Plugging the known; $$\boxed{ x=17.148\left( e^{-0.6887t} +0.6887t-1\right) }$$ c) You can use the trial and error method, or you can use any software calculator like WolframAlpha. In both cases the answer is $$t=\color{red}{\bf 9.9 }\;\rm s$$ Noting that in WolframAlpha, the two roots of time are $$t=9.918\;\rm s, \;-3.196\;s$$ The negative root is dismissed.
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.