Question

l Review What was Lewis's acceleration at t 4.20 s Careful measurements have been made of Olympic sprinters in the 100- Express your answer using four significant figures meter dash. A quite realistic model is that the sprinter's velocity is givern by a(1-e) where t is in s·L4 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 11.81m/s and b 0.6887s1 im Submit PrewausAnswe Request Ans鯉! Incorrect; Try Again; 9 attempts remaining ▼ Part D Find an expression for the distance traveled at time t. Express your answer in terms of the variables a, b, and t r(t)- Submit Previous Answers Request Answer

Answer 1

Answer :

1) if v(t) is the sprinter's velocity as a function of time with the given parameters a and b, then you just have to differentiate the velocity in respect to the time in order to obtain the sprinter's acceleration.

v(t) = a - a * exp(-bt)
a(t) = v'(t) = a * b * exp(-bt)

a(t) = 11.81 * 0.6887 e^{-(0.6887 * 4.2)} = 0.45 m/s²

2) if v(t) is the sprinter's velocity as a function of time with the given parameters a and b, then you just have to intigrate the velocity in respect to the time in order to obtain the sprinter's distance.

v(t) = a - a * exp(-bt)

$x(t) = \int v(t)$

x(t) = a² - a exp(-bt) / b

= 129.96 m