# Thread: Velocity of particle using graph? Physics?

1. ## Velocity of particle using graph? Physics?

I'm not sure how to figure this out:I'm not sure how to find how far the particle with using this graph? Any help? I took physics two years ago before college and I know I learned about this but I can't remember what to do, using either physics or calc methods. Thanks!

2. ## Re: Velocity of particle using graph? Physics?

$a(t) = 3t$

Acceleration is the derivative of velocity: $a(t) = v'(t)$. So:

$\displaystyle v(t) = \int a(t) dt = \dfrac{3}{2}t^2+C$

You are told initial velocity is 5. Initial velocity means t=0. So:

$v(0) = 5 = \dfrac{3}{2}(0)^2 + C$

shows $C=5$.

Hence, $v(t) = \dfrac{3}{2}t^2+5$.

Velocity is the derivative of position: $v(t) = s'(t)$. So, $\displaystyle s(t) = \int v(t)dt$.

Now, you are looking for the distance the particle has moved. That is just $\displaystyle s(3)-s(0) = \int_0^3 v(t)dt$:

$\displaystyle \int_0^3 v(t)dt = \int_0^3 \left(\dfrac{3}{2}t^2+5\right)dt = \left[\dfrac{1}{2}t^3 + 5t\right]_0^3 = 19.5$

3. ## Re: Velocity of particle using graph? Physics?

Originally Posted by SlipEternal
$a(t) = 3t$

Acceleration is the derivative of velocity: $a(t) = v'(t)$. So:

$\displaystyle v(t) = \int a(t) dt = \dfrac{3}{2}t^2+C$

You are told initial velocity is 5. Initial velocity means t=0. So:

$v(0) = 5 = \dfrac{3}{2}(0)^2 + C$

shows $C=5$.

Hence, $v(t) = \dfrac{3}{2}t^2+5$.

Velocity is the derivative of position: $v(t) = s'(t)$. So, $\displaystyle s(t) = \int v(t)dt$.

Now, you are looking for the distance the particle has moved. That is just $\displaystyle s(3)-s(0) = \int_0^3 v(t)dt$:

$\displaystyle \int_0^3 v(t)dt = \int_0^3 \left(\dfrac{3}{2}t^2+5\right)dt = \left[\dfrac{1}{2}t^3 + 5t\right]_0^3 = 19.5$
Thanks for showing the steps, I couldn't put that together because I was so focused on it being a physics problem I didn't even realize I could apply these steps.

4. ## Re: Velocity of particle using graph? Physics?

Physics involves a lot of applied math.