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.

  • 1
    $\begingroup$ Hello Tushar Mathur and welcome to Physics.SE. Here, we support only conceptual questions and we discourage questions that ask other users to solve specific homeworks.. Please take a look over our homework policy on how to ask homeworks..! $\endgroup$ Sep 30, 2012 at 17:07
  • $\begingroup$ Hi, I am new to this forum so was not really sure of the type of questions. Conceptually, here I just wanted to prove that if you are given the engine performance you can basically not find out the distance traveled by it at any point in time. $\endgroup$
    – tusharmath
    Oct 1, 2012 at 12:19
  • $\begingroup$ Converted it to a conceptual question. $\endgroup$
    – tusharmath
    Oct 1, 2012 at 12:23
  • $\begingroup$ the work made by the force is the variation of the kinetic energy $ W= \frac{m(v^{2}-v_{0}^{2})}{2} $ we still need to evaluate the initial speed , since we assume that if car stops final speed is $ v=0 $ the work done (if the force is constant) sis about $ W=F.x $ and x is the distance travelled $\endgroup$ Oct 1, 2012 at 12:35
  • $\begingroup$ @TusharMathur: Hi there Tushar. BTW, AGREED $\endgroup$ Oct 1, 2012 at 12:50

3 Answers 3


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:

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

  • $\begingroup$ awesome! I the "looking closely" part was the key to solving the problem. Thanks! $\endgroup$
    – tusharmath
    Oct 1, 2012 at 14:03

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 $\Delta t$.

  • $\begingroup$ Assuming that the initial speed is 0, when you will integrate the above equation once you will get what I have posted in my question. Which will ultimately leave you with a velocity parameter thus you will not be able figure out what is the actual distance traveled until u know the velocity. $\endgroup$
    – tusharmath
    Oct 1, 2012 at 13:40
  • $\begingroup$ @ja72, your derivation using $a(v)$ is nice and elegant! No need to go through two steps like I did, when you can get displacement directly from acceleration as a function of velocity. $\endgroup$
    – user11878
    Oct 1, 2012 at 13:42
  • 1
    $\begingroup$ @TusharMathur: divide by the left hand side and then integrate $\frac{m\frac{dv}{dt}}{F-uv}$ with respect to time. You should find that the velocity is an exponential in time that will limit off to a terminal velocity. You can then integrate THAT to get the position. $\endgroup$ Oct 1, 2012 at 13:57
  • $\begingroup$ @JerrySchirmer, yes you are right, I blame my poor integration skills. Thanks a lot for your help. $\endgroup$
    – tusharmath
    Oct 1, 2012 at 15:06

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.


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