# Integrating velocity function

1. Nov 18, 2012

1. The problem statement, all variables and given/known data

if $v = 8t - 3t^2$ and the body is at P

find where the body is from P when t=3

3. The attempt at a solution

I am very new to calculus and have just been taught basics of differentiation and integration.

I know that when you integrate there is an arbitrary constant so I have go this far (below).

$∫ ( v = 8t - 3t^2 ) dt = 4t^2 - t^3 + C$

I am not sure what to do with the constant, is it ignored and thus the answer is as below

$s = 4t^2 - t^3$
$s = (4 \times 9) - 27$
$s = 36 - 27$
$s = 9$metres ?

Or do I have to do something with the constant?

Last edited: Nov 18, 2012
2. Nov 18, 2012

### rock.freak667

Normally at t=0, s=0 and v=0. So your constant C will work out to be 0. So what you did is correct.

3. Nov 18, 2012

### Ray Vickson

The distance travelled is s(t) - s(0), from which C drops out.

RGV

4. Nov 19, 2012