Respuesta :
Answer:
a.
[tex]\displaystyle a(0 )=8.133\ m/s^2[/tex]
[tex]\displaystyle a(2)=2.05\ m/s^2[/tex]
[tex]\displaystyle a(4)=0.52\ m/s^2[/tex]
b.[tex]\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15[/tex]
c. [tex]t=9.9 \ sec[/tex]
Explanation:
Modeling With Functions
Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by
[tex]\displaystyle V(t)=a(1-e^{bt})[/tex]
For Carl Lewis's run at the 1987 World Championships, the values of a and b are
[tex]\displaystyle a=11.81\ ,\ b=-0.6887[/tex]
Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is
[tex]\displaystyle V(t)=11.81(1-e^{-0.6887t})[/tex]
a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?
To compute the accelerations, we must find the function for a as the derivative of v
[tex]\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})[/tex]
[tex]\displaystyle a(t)=8.133547\ e^{-0.6887t}[/tex]
For t=0
[tex]\displaystyle a(0)=8.133547\ e^o[/tex]
[tex]\displaystyle a(0 )=8.133\ m/s^2[/tex]
For t=2
[tex]\displaystyle a(2)=8.133547\ e^{-0.6887\times 2}[/tex]
[tex]\displaystyle a(2)=2.05\ m/s^2[/tex]
[tex]\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}[/tex]
[tex]\displaystyle a(4)=0.52\ m/s^2[/tex]
b. Find an expression for the distance traveled at time t.
The distance is the integral of the velocity, thus
[tex]\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C[/tex]
[tex]\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C[/tex]
To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates
[tex]\displaystyle x(0)=0=>11.81\times1.45201+C=0[/tex]
Solving for C
[tex]\displaystyle c=-17.1482\approx -17.15[/tex]
Now we complete the equation for the distance
[tex]\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15[/tex]
c. Find the time Lewis needed to sprint 100.0 m.
The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.
We are required to find the time at which the distance is 100 m, thus
[tex]\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100[/tex]
Rearranging
[tex]\displaystyle t+1.45\ e^{-0.6887t}=9.92[/tex]
We define an auxiliary function f(t) to help us find the value of t.
[tex]\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92[/tex]
Let's try for t=9 sec
[tex]\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92[/tex]
Now with t=9.9 sec
[tex]\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184[/tex]
That was a real close guess. One more to be sure for t=10 sec
[tex]\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081[/tex]
The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).
[tex]At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}[/tex]
