Is it possible to find out the distance traveled by a car if the force applied on it is given?

Question

Say you have car which produces $F$ amount of force which is transferred to the wheels directly.

Now assuming that there is air friction which is causing a retarding force proportional to the velocity of the car. I need to calculate the distance that is traveled by the car at any particular instance of time.

What I have done so far -

$$x = \frac{(F\cdot t - m\cdot v)}{u}$$

where: $x$ - distance traveled, $F$ - Force supplied to the wheels, $m$ - mass of the car, $v$ - velocity of the car, $t$ - time, $u$ - coefficient of air friction.

As you can see unless i know the velocity of the car, I can not tell the distance that was traveled.

Which proves that we can never determine the distance traveled by the car if we only know the force applied by the engine.

Answer 1

The answer to your title question is "of course it is possible!".

Conceptually, the resultant force acting on the car must equal the rate of change of the car's momentum:

$F_{net} = \dfrac{d}{dt}(mv) = m \dfrac{dv}{dt}$

The car's displacement, as a function of time, is found by integrating this equation twice so there will be two integration constants: the initial velocity and the initial position.

For a constant (net) force on the car, we get the familiar:

$x(t) = x_0 + v_0 t + \dfrac{F_{net}}{2m}t^2$

However, if the (net) force is not constant, the displacement function might be much more complicated.

For example, consider a net force that is part constant and part speed dependent:

$F_{net} = F - kv = m\dfrac{dv}{dt}$

Looking closely at this, note that there is a terminal velocity where $F - kv_{term} = 0$:

$v_{term} = \dfrac{F}{k}$

Substituting this into the previous equation and rearranging gives:

$\dfrac{1}{v_{term} - v}dv = \dfrac{k}{m}dt$

Integrating both sides and tidying up gives:

$v(t) = (v_0 - v_{term})e^{\frac{-kt}{m}} + v_{term}$

To find $x(t)$, integrate $v(t)$ above:

$x(t) = \dfrac{m}{k}(v_0 - v_{term})(1 - e^{\frac{-kt}{m}}) + v_{term}\cdot t + x_0$

Answer 2

I don't agree with Tushar; the way the problem was solved in the question is incorrect. If the engine produces force $F$ (which was assumed constant, if I understood correctly - quite some electric motor there), the equation of motion will be $$ \sum F = F - uv = ma = m \frac{dv}{dt} $$ which is a separable differential equation. Easy to solve for $v(t)$, if $F$ is constant. However, if you don't know the initial speed $v_0$, you will only obtain the change in speed $\Delta v$ as a function of time interval $\Delta t$. This should be straightforward to integrate, and thereby to obtain displacement, or change in position $\Delta x$ as a function of time interval $\Delta t$.

Answer 3

If you know the acceleration at any time as a function of speed $a(v)$ then the time and distance traveled between an initial and final speed is:

$$ \Delta x = \int_{v_1}^{v_2} \frac{v}{a(v)}\,{\rm d} v $$ $$ \Delta t = \int_{v_1}^{v_2} \frac{1}{a(v)}\,{\rm d} v $$

Why?? $$ a {\rm d}x = \frac{{\rm d}v}{{\rm d}t} {\rm d}x = v {\rm d} v$$ $$ {\rm d}x = \frac{v}{a} {\rm d} v $$ $$ \int {\rm d}x = \int \frac{v}{a} {\rm d} v $$ and $$ {\rm d} v = a {\rm d}t$$ $$ {\rm d}t = \frac{1}{a} {\rm d} v $$ $$ \int {\rm d}t = \int \frac{1}{a} {\rm d} v $$

So now what you have to do is devise the formula for $a(v) = \frac{F-k v}{m} $ and do the above integrals.