In a women\'s 100-m race, accelerating uniformly, Laura takes 1.87 s and Healan

Question

In a women's 100-m race, accelerating uniformly, Laura takes 1.87 s and Healan 3.12 s to attain their maximum speeds, which they each maintain for the rest of the race. They cross the finish line simultaneously, both setting a world record of 10.4 s. (a) What is the acceleration of each sprinter? aLaura = m/s2 aHealan = m/s2 (b) What are their respective maximum speeds vLaura,max = m/s vHealan,max = m/s (c) Which sprinter is ahead at the 6.20-s mark, and by how much? is ahead by m. (d) What is the maximum distance by which Healan is behind Laura? m At what time does that occur? s

Explanation / Answer

I can suggest a couple of ways of getting d). First, write out the equations of motion for both runners, meaning write out expressions for x(t) for each runner. Now, construct the quantity x2(t)-x1(t) where x2 and x1 are the time dependent positions of each runner take d/dt of x2(t)-x1(t) and set that derivative equal to zero; the value of t that satisfies this equation will be the time of maximum separation; substitute that value into x2(t)-x1(t) to find the actual separation second, you could plot x2(t) and x1(t) on the same set of axes and visually determine the time and amount of separation hope this helps note added later...I checked your work and agree with your accelerations; after looking into it a bit more, I think the best way to solve this is graphically (or a combination of graphing and calc) if you plot x1(t)-x2(t), you will see that the maximum occurs around 2.8 s and has a magnitude of around 4.4m this is at a time when the first runner has reached max speed, and the second runner is still accelerating, so we can write the separation function as: x1(t)-x2(t) = 10.64+10.64(t-2) - 1.875 t^2 differentiate this expression and find that d/dt(x1(t)-x2(t))=0 when t=2.83s, so this is the time of max separation; now, substitute this into the x1(t)-x2(t) expression and find the value of the separation (I get 4.45m)

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