Careful measurements have been made of Olympic sprinters in the 100-meter dash.

Question

Careful measurements have been made of Olympic sprinters in the 100-meter dash. A quite realistic model is that the sprinter's velocity is given by v_x = a(1 - e^-bt) where t is in s, v_x is in m/s, and the constants a and b are characteristic of the sprinter. Sprinter Carl Lewis's run at the 1987 World Championships is modeled with a = 11.81 m/s and b = 0.6887 s^-1. a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s? b. Find an expression for the distance traveled at time t. c. Your expression from part b is a transcendental equation, meaning that you can't solve it for t. However, it's not hard to use trial and error to find the time needed to travel a specific distance. To the nearest 0.01 s, find the time Lewis needed to sprint 100.0 m. His official time was 0.01 s more than your

Explanation / Answer

2.

a)

Vx = a (1 - e-bt)

a = 11.81 , b = 0.6887

so Vx = (11.81) (1 - e-(0.6887)t)

acceleration is given as

ax = dVx/dt = d/dt ((11.81) (1 - e-(0.6887)t)) = (11.81) (0.6887) e-(0.6887)t

at t = 0

ax = (11.81) (0.6887) e-(0.6887)t = (11.81) (0.6887) e-(0.6887)(0) = 8.13 m/s2

at t = 2

ax = (11.81) (0.6887) e-(0.6887)t = (11.81) (0.6887) e-(0.6887)(2) = 2.05 m/s2

at t = 4

ax = (11.81) (0.6887) e-(0.6887)t = (11.81) (0.6887) e-(0.6887)(4) = 0.52 m/s2

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