- #1
lorenz0
- 151
- 28
- Homework Statement
- In a 100m race, two runners run the first 40m both in ##t_1=5s##. Runner A runs with a constant and maximal acceleration ##a_A## for the first ##2.8s## and then runs at constant speed ##v_A##. The second runner B runs at a constant and maximal acceleration ##a_B## for the first ##3.2s## and then he also runs at a constant speed ##v_B##. Determine:
1) who wins the race and the time he takes to do so;
2) how much time runner B would take to run the 100m if, after having achieved the maximum speed ##v_B##, started decelerating with ##a=-kv##, where ##k=0.05 s^{-1}##
- Relevant Equations
- ##x(t)=x_0+v_0 t+\frac{1}{2}at^2, v(t)=v_0+at, a=\frac{dv}{dt}, v=\frac{dx}{dt}##
1. $$x_{A}(t)=\begin{cases}\frac{1}{2}a_A t^2 & 0\leq t\leq 2.8\\ \frac{1}{2}a_A 2.8^2 +v_A t & t>2.8\end{cases}=\begin{cases}\frac{1}{2}a_A t^2 & 0\leq t\leq 2.8\\ \frac{1}{2}a_A 2.8^2 +a_A\cdot 2.8\cdot t & t>2.8\end{cases}$$ so ##x_A(5)=40## implies ##a_A=\frac{40}{17.92}m/s##. Similarly for B we have $$x_{B}(t)=\begin{cases}\frac{1}{2}a_B t^2 & 0\leq t\leq 3.2\\ \frac{1}{2}a_B 3.2^2 +v_B t & t>3.2\end{cases}=\begin{cases}\frac{1}{2}a_B t^2 & 0\leq t\leq 3.2\\ \frac{1}{2}a_B 3.2^2 +a_B\cdot 3.2\cdot t & t>3.2\end{cases}$$ so ##x_B(5)=40## implies ##a_B=\frac{40}{21.12}m/s##. Thus the time ##t_{100A}## that it takes for A to reach 100 m is $$100=\frac{1}{2}a_A 2.8^2 +a_A\cdot 2.8\cdot t_{100A}\Rightarrow t_{100A}=\frac{1}{2}(\frac{200}{2.8 a_A}-2.8)=14.6s$$. Similarly, the time ##t_{100B}## that it takes for B to reach 100m is $$100=\frac{1}{2}a_B\cdot 3.2^2+a_B\cdot 3.2\cdot t_{100B}\Rightarrow t_{100B}=\frac{1}{2}(\frac{200}{3.2a_B}-3.2)=14.9s$$ so A reaches the 100m before B does.
2. $$a=\frac{dv}{dt}=-kv\Rightarrow \frac{dv}{v}=-kdt\Rightarrow \int_{v_B}^{v(t)}\frac{dv'}{v'}=-k\int_{3.2}^{t}dt' \Rightarrow v(t)=v_B e^{3.2k-kt}=\frac{dx}{dt}\Rightarrow \int_{40}^{100}dx=\int_{3.2}^{t_{100}}v_B e^{3.2k-kt}dt\Rightarrow 60=a_B\cdot 3.2e^{3.2k}\int_{3.2}^{t_{100}}e^{-kt}dt\Rightarrow t_{100}=-20\cdot\ln(\frac{101}{200}e^{-\frac{3.2}{20}})\approx 16.864 s$$
2. $$a=\frac{dv}{dt}=-kv\Rightarrow \frac{dv}{v}=-kdt\Rightarrow \int_{v_B}^{v(t)}\frac{dv'}{v'}=-k\int_{3.2}^{t}dt' \Rightarrow v(t)=v_B e^{3.2k-kt}=\frac{dx}{dt}\Rightarrow \int_{40}^{100}dx=\int_{3.2}^{t_{100}}v_B e^{3.2k-kt}dt\Rightarrow 60=a_B\cdot 3.2e^{3.2k}\int_{3.2}^{t_{100}}e^{-kt}dt\Rightarrow t_{100}=-20\cdot\ln(\frac{101}{200}e^{-\frac{3.2}{20}})\approx 16.864 s$$
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