# Solve for velocity, if acceleration is a function of velocity?

Yes, this is a canned question, because canned questions are simply solvable to understand ideas. No, this is not homework.

For simplicity, let's assign values.

A given object weights 1000 kg, it only has one force acting on it, and that's friction: $F_{t} = -200*v(t)$.

Let's say, after 5 seconds, what is its final velocity?

In case you haven't brushed up on your physics lately, here's a rundown of my mathematical problem.

F = ma, so a = F/m. With the numbers above, that means $a = \dfrac{-v}{5}$.

If I want to know final velocity after 5 seconds, it will basically be

$$30 - \int_0^5 -v/5 \,dt= ??$$ ..two different variables.

I believe the answer should involve ln (or e), but I can't seem to set it up correctly.

• what's a "canned question" ? – Sled Apr 15 '15 at 20:39
• Question should probably also provide the initial velocity. The answers appear to arbitrarily be using an initial velocity of 30mps. – Mooing Duck Apr 15 '15 at 22:45

The acceleration of an object is the derivative of its velocity. The force acting on the object produces an acceleration according to the usual rule $F = ma$. The equation you get is $$1000 v'(t) = -200 v(t).$$ In other words, $v'(t) = - \frac 15 v(t)$ which has the general solution $v(t) = v(0)e^{-\frac 15 t}$.

If $v(0) = 30$, you get $v(5) = \frac{30}{e}$.

• +1 though that might be easier to read as $v(t)=v(0)\exp\left(-\dfrac15 t\right)$ or $v(t)=v(0)e^{-t/5}$ – Henry Apr 15 '15 at 17:20

Oh wow!

The answer just came to me!

I'm not sure if this will help anyone, but I figure it's probably better to just provide the answer for posterity.

So, really what we're talking about here is:

$$\frac{dv}{dt} = \frac{-v}{5}$$

So, $$\frac{dv}{v} = \frac{-1}{5}dt$$

Then integrate both sides to get

$$ln(v) = \frac{-1}{5}t + C$$

which rearranges to

$$v = C*e^{-t/5}$$

And C can be solved with the initial condition of t = 0, v = 30. Which means C = 30.

So, our equation for velocity at any time becomes

$$v = 30e^{-t/5}$$

• You indeed did get it! +1. – Robert Lewis Apr 15 '15 at 17:39

Recall the definition of $a$:

$a = \dfrac{dv}{dt}; \tag{1}$

thus, with the numbers provided,

$\dfrac{dv}{dt} = -\dfrac{1}{5}v; \tag{2}$

this yields

$v(t) = v(t_0)e^{-(t - t_0) / 5}; \tag{3}$

here I have supplied the number not explicitly provided, which is $v(t_0)$, the initial velocity at time $t = t_0$; apparently $v(t_0) = 30 M/sec$, assuming we measure distance in meters (M). Plugging this and $t = t_0 + 5 sec$ into (3) yields

$v(t_0 + 5) = 30e^{-1}M/sec = \dfrac{30}{e}M/sec. \tag{4}$

The velocity decreases by a factor of $e$.

• Ha, you answered it while I was writing out the one I realized. :) – Meshach Apr 15 '15 at 17:38
• Yuppers, a tie at the finish line! – Robert Lewis Apr 15 '15 at 17:40
• @RobertLewis HEY,YUPPERS IS MY LINE! lol – Mathemagician1234 Apr 15 '15 at 17:56