The force acting on a car with mass $m$ and travelling in a straight line along the $x$-axis with velocity $v > 0$ is:

$$F = m(2a^2k − 3akv − 2kv^2)$$

where $a>$ and $k>0$ are such that $a,k \in \mathbb{R}$.

If the car is at rest at $t = 0$, determine the velocity of the car as a function of time.

Show that $$v \rightarrow \frac{a}{2} \text{ as }t \rightarrow +\infty$$

Hey! I am still a newbie with mechanics/physics and I am having a really hard time understanding and solving such questions. Any help? Thank you in advance.


1 Answer 1


You have the force

$$F(v) = 2ma^2k - 3makv - 2mkv^2 \implies \frac{\mathrm dv}{\mathrm dt} = 2a^2k - 3akv - 2kv^2$$

Now let's call $A:= 2k$, $B := 3ak$ and $C:=-2a^2k$ so you will have to solve

$$-\frac{\mathrm dv}{\mathrm dt} = Av^2 + Bv + C \implies \frac{\mathrm dv}{Av^2+Bv+C} = -\mathrm dt$$

Now we define

$$v_{\pm} = \frac{-B \pm \sqrt{B^2-4AC}}{2A}$$

and then we will have that

$$\int_{0}^v\frac{\mathrm dv'}{(v'-v_-)(v' - v_+)} = -t$$

But we can use that

$$\frac{1}{(v-v_-)(v - v_+)} = \frac{1}{(v_- - v_+)(v - v_-)} - \frac{1}{(v_- -v_+)(v - v_+)}$$

Using the values we note that $v_+ = a/2$ and $v_- = -2a$ and $v_- - v_+ = -5a/2$. So

$$\frac{1}{(v-v_-)(v - v_+)} = \frac{1}{(v_- - v_+)(v - v_-)} - \frac{1}{(v_- -v_+)(v - v_+)} = \frac{-2}{5a}\frac{1}{(v + 2a)} - \frac{-2}{5a}\frac{1}{(v - a/2)} = \frac{-2}{5a}\left \{\frac{1}{v+2a} - \frac{1}{v - a/2} \right\}$$

So we have to solve

$$\frac{-2}{5a}\int_0^v\frac{\mathrm dv'}{v' + 2a} = \frac{-2}{5a}\ln\left(\frac{v+2a}{2a}\right)$$

$$\frac{-2}{5a}\int_0^v\frac{\mathrm dv}{v' - a/2} = \frac{-2}{5a}\ln\left(\frac{v - a/2}{-a/2}\right)$$


$$\ln\left(\frac{(v-a/2)(2a)}{(-a/2)(v+2a)}\right) = -\frac{5at}{2}$$

Which will give the function $v = v(t)$

$$v(t) = \frac{2a(1 - e^{-5at/2})}{4 + e^{-5at/2}}$$

Now take the limit where $t \to \infty$ then you have

$$\lim_{t \to \infty}v(t) = \frac{2a}{4} = \frac{a}{2}$$

As the exercise was asking. This is a very important type of exercise in elementary Newtonian mechanics. This force has two orders of an friction ans you should in this type of exercises solve an differential equation (usually they are non-linear ).


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