# Short calculus task

## Homework Statement

A particle initially moving at 5m/s has an acceleration of a=-.5v.
Find v(t)

## The Attempt at a Solution

i know the anti derivative of acceleration is -(v^2)/4
but i have no idea how to put velocity in terms of time when acceleration in terms of v is given

## Answers and Replies

It's a separable differential equation:

$$a = \frac{dv}{dt} = -\frac{1}{2}v$$

Does that help?

not really...i still don't fully understand how to put v as a function of time

When solving a separable differential equation, you treat the derivative (dv/dt) as a fraction. Get all the v's and dv's on one side and all the t's and dt's on the other side:
dv/dt=-1/2 v
1/v dv=-1/2 dt

Then integrate both sides and solve for v.

the answer given in the notebook is "v=5 exp(-.5t)"...i don't understand :s
when i integrate i get
ln|v| = -1/2t
solving for v = e^(-.5t)
where is this 5 coming from??

Last edited:
You forgot the constant in your integration. What you have is an initial condition problem: v(0)=5. So when you integrate both sides, you get
ln|v| = -1/2 t + c
v=e^(-1/2 t+c)

It makes things neater to get that constant out of the exponent, using the laws of exponents:
v=e^c*e^(-1/2 t)

Now, since c is a constant, e^c is a constant, so we can just cal lthat a new constant to make it neater:
Let k=e^c
v=k*e^(-1/2 t)

With your initial condition v(0)=5:
5=k*e^(-1/2*0)
5=k*1
k=5

So v=5e^(-1/2 t)

ahh i c thx alot!