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

answer, showing that this model is very good, but not perfect.

Short answer

Acceleration is defined as the rate of change of velocity with respect to time.

Step-by-step solution

Calculating the acceleration :

The given velocity is $v_x=a\left(1-e^{-bt}\right)$

Acceleration is calculated as the derivative of velocity with respect to time.

$$\begin{gathered} \overrightarrow{a_x}=\frac{d{v}_x}{dt} \\ =\frac{d}{dt}\left(a(1-e^{-bt}\right)) \\ =0-\left(a.-b.e^{-bt}\right) \\ =ab.e^{-bt} \end{gathered}$$

Calculating the acceleration at t=0.00s:

The acceleration is calculated above and substitute a=11.81 m/s and b=0.6887s^-1 :

$$\begin{gathered} \overrightarrow{a_x}=11.81m/s\times 0.6887s^{-1}\times e^{-0.6887t} \\ =8.31m/s^2\times e^{-0.6887t} \end{gathered}$$

At t=0.00s, the value of acceleration is calculated as:

$$\begin{gathered} \overrightarrow{a_{x0}}=8.13\times e^{-0.6887\times 0} \\ =8.13m/s^2 \end{gathered}$$

Calculating acceleration at t=2.00s and t=4.00 s:

At t=2.00s, the value of acceleration is calculated as:

$$\begin{gathered} \overrightarrow{a_{x2}}=8.13\times e^{-0.6887\times 2} \\ =2.05m/s^2 \end{gathered}$$

At t=4.00s, the value of acceleration is calculated as :

$$\begin{gathered} \overrightarrow{a_{x4}}=8.13\times e^{-0.6887\times 4} \\ =8.13\times 0.0636 \\ =0.52m/s^2 \end{gathered}$$

Calculating the distance travelled by the sprinter at time t:

Distance is the anti derivative of velocity with respect to time.

$$\begin{gathered} \overrightarrow{r}=\int v_{x}dt \\ =\int a(1-e^{-bt})dt \\ =at+\frac{a}{b}{e}^{-bt}+C \end{gathered}$$

Calculating the integrating constant C: at t=0, r=0

$$\begin{gathered} \overrightarrow{r}=a.0+\frac{a}{b}{e}^{-b\times 0}+C \\ 0=\frac{a}{b}+C \\ C=-\frac{a}{b} \end{gathered}$$

Therefore, the distance travelled at t is given by:

$$\begin{gathered} \overrightarrow{r}=at+\frac{a}{b}\times e^{-bt}-\frac{a}{b} \\ =at+\frac{a}{b}(e^{-bt}-1) \end{gathered}$$

Calculating the time needed to sprint 100.0 m:

Since the equation of r is a transcendental equation, assume that the average time of the sprinter for 100 m dash is 10 s.

Substitute t=10s into the above equation for r:

$$\begin{gathered} \overrightarrow{r}=11.81\times 10.00+\frac{11.81}{0.6887}(e^{-0.6887\times 10.00}-1) \\ =100.97m \end{gathered}$$

Repeat the process of hit and trial method and substitute at t=9.92 s:

$$\begin{gathered} \overrightarrow{r}=11.81\times 9.92+\frac{11.81}{0.6887}(e^{-0.6887\times 9.92}-1) \\ =100.03m \end{gathered}$$

Therefore, at t=9.92 s, the length of the dash is 100m.

Therefore, the time Lewis needed to sprint 100.0 m is 9.92 s.