[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:
It is real, inverted, and smaller than the object.
Explanation:
First of all, we can use the lens equation to find the location of the image:
where
q is the distance of the image from the lens
f = 15 cm is the focal length (positive for a converging lens)
p = 50 cm is the distance of the object from the lens
Solving the equation for q,
The distance of the image from the lens is positive, so we can already conclude that the image is real.
Now we can also write the magnification equation:
where are the size of the image and of the object, respectively.
Substituting p = 50 cm and q = 21.3 cm, we have
So from this relationship we observe that:
--> this means that the image is smaller than the object, and
--> this means that the image is inverted
so, the correct answer is
It is real, inverted, and smaller than the object.