Kinematics: two runners with different and non-constant accelerations

AI Thread Summary
The discussion focuses on the kinematics of two runners, A and B, with different non-constant accelerations. Calculations show that runner A reaches 100 meters in approximately 14.6 seconds, while runner B takes about 14.9 seconds, indicating A is faster. Both runners cover 40 meters in 5 seconds but must run the subsequent 60 meters at a higher average speed than the first segment. There is a critique regarding the calculation of accelerations for both runners, suggesting that the time variables were not correctly applied in the velocity equations. Overall, the analysis highlights discrepancies in the initial calculations of acceleration and time taken to complete the race distance.
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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. $$a=\frac{dv}{dt}=-kv\Rightarrow \frac{dv}{v}=-kdt\Rightarrow \int_{v_B}^{v(t)}\frac{dv'}{v'}=-k\int_{3.2}^{t}dt' \Rightarrow v(t)=v_B e^{3.2k-kt}=\frac{dx}{dt}\Rightarrow \int_{40}^{100}dx=\int_{3.2}^{t_{100}}v_B e^{3.2k-kt}dt\Rightarrow 60=a_B\cdot 3.2e^{3.2k}\int_{3.2}^{t_{100}}e^{-kt}dt\Rightarrow t_{100}=-20\cdot\ln(\frac{101}{200}e^{-\frac{3.2}{20}})\approx 16.864 s$$
 
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They both run 40m in 5s. They must run the second 60m at a greater average speed than the first 40m. Therefore, both should take less than 12.5s, and not nearly 15s.

Your answers to 1) can't be right.
 
##a_A## and ##a_B## didn't come out right because you didn't use ##t_v=5-t_a## in the ##v## terms
 
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