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A car of mass m, initially at rest at time t = 0, is driven to
theright, as shown above, along a straight,
horizontal road with the engine causing a constant force Fo to
beapplied. While moving, the car encounters a
resistance force equal to -kv, where v is the velocity of the
carand k is a positive constant.
b. Determine the horizontal acceleration of the car in terms of
k,v, Fo, and m.
c. Derive the equation expressing the velocity of the car as
afunction of time t in terms of k, Fo, and m.

Answer :

Answer:

[tex]a=\dfrac{F_0-kv}{m}[/tex]

[tex]v=\dfrac{F_0}{k} (1-e^{\dfrac{-kt}{m}})[/tex]

Explanation:

Given that

Constant force =Fo

mass = m

F= - k v

We know that

F(net)= m a

a=acceleration

Fo - k v = m a

[tex]a=\dfrac{F_0-kv}{m}[/tex]

The acceleration in terms k  Fo and m given as

[tex]a=\dfrac{F_0-kv}{m}[/tex]

We also know that

[tex]a=\dfrac{dv}{dt}[/tex]

[tex]\dfrac{dv}{dt}=\dfrac{F_0-kv}{m}[/tex]

[tex]\dfrac{dv}{F_0-kv}=\dfrac{dt}{m}[/tex]

[tex]\int_{0}^{v}\dfrac{dv}{F_0-kv}=\int_{0}^{t}\dfrac{dt}{m}\\\left[\dfrac{-1}{k}\ln(F_0-kv)\right]_0^v=\left[\dfrac{t}{m}\right]_0^t\\\dfrac{t}{m}=\dfrac{-1}{k}\ln\dfrac{F_0-kv}{F_0}[/tex]

[tex]\dfrac{F_0-kv}{F_0}=e^{\dfrac{-kt}{m}}\\\\F_0-kv=F_oe^{\dfrac{-kt}{m}}\\kv=F_0-F_oe^{\dfrac{-kt}{m}}[/tex]

[tex]v=\dfrac{F_0}{k} (1-e^{\dfrac{-kt}{m}})[/tex]

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