[7 pts] 1. Suppose that a car starts from rest, its engine providing an acceleration of 4 ft/s2 while air resistance provides 0.
1 ft/s2 of deceleration for each ft per s (foot per second) of the car’s velocity. Set up an Initial Value Problem (IVP) for the velocity of the car. Solve the IVP and use the solution to find the car’s limiting velocity. If the car starts at position x(0) = x0 = 0, ft, what position will it be at after 1 2 hour? Be sure to set up your coordinate system and include units in all of your answers.
1 answer:
Answer:
a) Initial Value Problem
dv/dt = 4 - 0.1v
v(0) = 0
b) solution to the IVP
v(t) = 40(1 - e^(-t/10))
c) Limiting velocity
Vo = 40 ft/s
Position of the car after 12 hours
X = 14,390 ft
Explanations:
The complete explanations of each of the sections contained in the question are in the files attached to this solution.
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Answer:
b. Friction decreased when he went from pavement to ice and then increased two more times.
Explanation:
Frictional force depends on the normal force of the surface and a friction coefficient.
Since we're talking about the same car, the value of will remain constant whereas μ will represent the change in the frictional coefficient of the surface. Now we consider the different surfaces, cars will slide in an icy road which means that the frictional coefficient is smaller than the pavement.
After Joshua returns to the pavement road, the resulting frictional force increases and will do so one more time when he reaches the gravel road. Gravel roads have greater frictional coefficients than pavement roads which means the frictional force will increase a second time.