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: 79

Answer

a) $10\;\rm s$ b) $3.83\;\rm m/s^2$ c) $6.4\%$

Work Step by Step

a) To find the race time, we need to find the distance during the acceleration. $$x_1=x_0+v_{0x}t_1+\frac{1}{2}a_xt_1^2=0+0+\left(\frac{1}{2}\cdot 3.6\cdot\left[\dfrac{10}{3}\right]^2 \right)$$ $$x_1=20\;\rm m\tag 1$$ Thus, the rest distance will be at a constant speed which is given by $$v_{x1}=v_{0x}+a_xt_1=0+3.6\cdot \dfrac{10}{3} $$ $$v_{x1}=12\;\rm m/s\tag 2$$ This is the speed the sprinter completes the race at. $$\Delta x=100-x_1=v_{x1}t_2$$ Solving for $t_2$; $$t_2=\dfrac{100-x_1}{v_{x1}}$$ Thus the total time is given by $$t=t_1+t_2=\dfrac{10}{3}+\dfrac{100-x_1}{v_{x1}}$$ Plugging from (1) and (2); $$t= \dfrac{10}{3}+\dfrac{100-20}{12}$$ $$t=\color{red}{\bf10}\;\rm s$$ b) The whole distance is given by $$x=x_1+v_{x1}t_2$$ $$100=x_1+12t_2$$ where $x_1=\frac{1}{2}a_xt_1^2$ $$100=\frac{1}{2}a_xt_1^2+12t_2\tag 3$$ $$v_{x1}=v_0+a_xt_1$$ $$12=a_xt_1\tag{Plug into (3)}$$ $$100=\frac{1}{2}t_1 \;\overbrace{ (a_xt_1) }^{12} +12t_2 $$ $$100=6t_1+12t_2$$ Noting that $t_1+t_2=9.9$, thus, $t_2=9.9-t_1$ $$100=6t_1+12(9.9-t_1)$$ Hence, $$t_1=3.13\;\rm s$$ Thus, $$a_x=\dfrac{12}{3.13}$$ $$a_x=\color{red}{\bf3.83}\;\rm m/s^2$$ c) Decreasing his time by 1$\%$ means he will run the 100-m race in $$t=t_{original}-0.01t_{original}=10-(0.01\cdot 10)=\bf 9.9\;\rm s$$ which is the same time for the second case when his acceleration is 3.83 m/s$^2$. Thus, the percentage increase in his acceleration is given by $${ a_x\%}_{\rm increase}=\dfrac{a_{x2}-a_{x1}}{a_{x1}}\times100\%$$ $${ a_x\%}_{\rm increase}=\dfrac{3.83-3.6 }{3.6}\times100\%$$ $${ a_x\%}_{\rm increase}=\color{red}{\bf 6.4}\%$$
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