A car of mass \(\displaystyle 600 \ kg\) is driven along a straight road. The resistance to motion of the car is \(\displaystyle kv^2\) Newtons, where \(\displaystyle v \ m/s\) is the velocity of the car at time \(\displaystyle t\) seconds and \(\displaystyle k\) is a constant. The engine of the car is producing \(\displaystyle 8 \ kW\) of output power.
A.
i. Draw a free body diagram of the system.
ii. Derive the differential equation of motion of the vehicle, and describe it in the form:
\(\displaystyle \frac{dv}{dt} + Av^2  \frac{B}{v} = 0\)
Where \(\displaystyle A\) and \(\displaystyle B\) are constants to be found.
B. When the velocity of the car is \(\displaystyle 20 \ m/s\), the engine is turned off.
i. Draw a free body diagram of the system.
ii. Derive the differential equation of motion of the car, and describe it in the form:
\(\displaystyle \frac{dv}{dt} + Av^2 = 0\)
Where \(\displaystyle A\) is a constant to be found.
iii. Find, in term of \(\displaystyle k\), an expression for the velocity of the car. Make sure your expression is in the form \(\displaystyle v = f(t)\)
This is what I did for \(\displaystyle A\)
I am not sure if this is the correct free body diagram. Can anyone help me, please?
Now, I will try to derive the differential equation.
\(\displaystyle F_e  F_R = ma = m\frac{dv}{dt}\)
\(\displaystyle m\frac{dv}{dt} + kv^2  F_e = 0\)
Now I will use the output power
\(\displaystyle F_e = \frac{P}{v}\)
\(\displaystyle m\frac{dv}{dt} + kv^2  \frac{P}{v} = 0\)
\(\displaystyle \frac{dv}{dt} + \frac{kv^2}{m}  \frac{P}{mv} = 0\)
Is this a correct derivation? Please, let me know if I did something wrong in the derivation or in the free body diagram.
I also noticed that if I want to solve this differential equation with separable method, it gives a very complicated solution while wolfram can solve it easily \(\displaystyle v(t) = Atv^2 + \frac{Bt}{y} + c_1\), if I let \(\displaystyle A = \frac{k}{m}\) and \(\displaystyle B = \frac{P}{m}\)
Why is that?
A.
i. Draw a free body diagram of the system.
ii. Derive the differential equation of motion of the vehicle, and describe it in the form:
\(\displaystyle \frac{dv}{dt} + Av^2  \frac{B}{v} = 0\)
Where \(\displaystyle A\) and \(\displaystyle B\) are constants to be found.
B. When the velocity of the car is \(\displaystyle 20 \ m/s\), the engine is turned off.
i. Draw a free body diagram of the system.
ii. Derive the differential equation of motion of the car, and describe it in the form:
\(\displaystyle \frac{dv}{dt} + Av^2 = 0\)
Where \(\displaystyle A\) is a constant to be found.
iii. Find, in term of \(\displaystyle k\), an expression for the velocity of the car. Make sure your expression is in the form \(\displaystyle v = f(t)\)
This is what I did for \(\displaystyle A\)
I am not sure if this is the correct free body diagram. Can anyone help me, please?
Now, I will try to derive the differential equation.
\(\displaystyle F_e  F_R = ma = m\frac{dv}{dt}\)
\(\displaystyle m\frac{dv}{dt} + kv^2  F_e = 0\)
Now I will use the output power
\(\displaystyle F_e = \frac{P}{v}\)
\(\displaystyle m\frac{dv}{dt} + kv^2  \frac{P}{v} = 0\)
\(\displaystyle \frac{dv}{dt} + \frac{kv^2}{m}  \frac{P}{mv} = 0\)
Is this a correct derivation? Please, let me know if I did something wrong in the derivation or in the free body diagram.
I also noticed that if I want to solve this differential equation with separable method, it gives a very complicated solution while wolfram can solve it easily \(\displaystyle v(t) = Atv^2 + \frac{Bt}{y} + c_1\), if I let \(\displaystyle A = \frac{k}{m}\) and \(\displaystyle B = \frac{P}{m}\)
Why is that?
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