[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:
The minimum coefficient of friction required is 0.35.
Explanation:
The minimum coefficient of friction required to keep the crate from sliding can be found as follows:
Where:
μ: is the coefficient of friction
m: is the mass of the crate
g: is the gravity
a: is the acceleration of the truck
The acceleration of the truck can be found by using the following equation:
Where:
d: is the distance traveled = 46.1 m
: is the final speed of the truck = 0 (it stops)
: is the initial speed of the truck = 17.9 m/s
If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.
Therefore, the minimum coefficient of friction required is 0.35.
I hope it helps you!