Answer:
For 120 year (1051200 hours), we don´t know which function give us the velocity of the car because it is stated on the question that [tex]v(t)= 60te-t4[/tex] is only valid for the first 3 hours.
I assume you mistake 120 years with 120 minutes (2 hours). If this is the case, then the question can be solve.
Answer (For 120 minutes):
First, we need to have the same unit in for our variable and for the domain of the function. It can be done by change 120 to hours or changing t from 0 to 3 to minutes.
The easiest way is doing [tex]120 min * \frac{1h}{60min} = 2h[/tex]
Having the variable and the domain in the same unit, allow us to solve this problem as a kinematics problem.
You may know that velocity is equal to the varion of space over time.
mathematically, that means:
[tex]\frac{dx}{dt} = v(t)[/tex]. If we know the function v(t), we can obtain x(t) by solving this equality as a differential equation.
[tex]\frac{dx}{dt} = v(t)\\\\dx=v(t) dt\\\\\\\int\limits^A_B {} \, dx = \int\limits^a_b {v(t)} \, dt[/tex]
Before jumping into the integral, we need to determine the integration point A-B and a-b.
As the question ask for a the distain between two points, we can assume the first point is x0=0 and the last point is xf=d, where d represents the travelled distance. So A-B=0-d.
For a-b, the question ask for a 2 hours trip. So we adding the velocity from the hour zero to the hour 2. So a-b=0-2.
[tex]\\\int\limits^d_0 {} \, dx = \int\limits^2_0 {v(t)} \, dt\\\\ v(t)= 60te-t4\\\\\\\int\limits^d_0 {} \, dx = \int\limits^2_0 {60te-t4} \, dt\\\\\\d= \int\limits^2_0 {60te} \, dt + \int\limits^2_0 {-t4} \, dt\\d= 60e \int\limits^2_0 {t} \, dt - 4 \int\limits^2_0 {t} \, dt\\d= 60 e \frac{(2^{2}-0) }{2} -4 \frac{(2^{2}-0) }{2} = 60e*2 - 4*2=120e-8.[/tex]
So the travelled distance d is 318,19 miles.
