# Thread: Rectilinear Motion: finding displacement & distance traveled w/ integration (2nd ? )

1. ## Rectilinear Motion: finding displacement & distance traveled w/ integration (2nd ? )

v(t) = t^3 - 3t^2 +2t 0 < t <3

Alright, I was fine finding the displacement ...

I did sigma(t^3-3t^2 + 2t)dt

I got t^3-3t^2 + 2t and evaluated on 0-3

(81/4 - 108/4 + 36/4) = 9/4 meters

However I am having trouble finding the total distance traveled....

Since it is over the period of 0-3, I know I need to split up the integrals when the sign of v(t) changes |(t^3-3t^2 + 2t)dt|

Setting v(t) to 0 I get 0, 1, 2 ... The sign changes from negative to positive at t=1

I took sigma 0,1 |t^3-3t^2 + 2t dt|- sigma 1,2 |t^3-3t^2 + 2t dt|

Not sure what my mistake was... please help...I have a test on this tomorrow and I am so lost
The solutions manual says the answer is 11/4 meters

2. ## Re: Rectilinear Motion: finding displacement & distance traveled w/ integration (2nd

The velocity function may be factored as follows:

$\displaystyle v(t)=t(t-1)(t-2)$

So, we see there are roots at $\displaystyle t=0,\,1,\,2$. Testing the sign we find $\displaystyle v(t)<0$ on $\displaystyle (1,2)$ and so the total distance traveled is:

$\displaystyle D=\int_0^1 t^3-3t^2+2t\,dt-\int_1^2 t^3-3t^2+2t\,dt+\int_2^3 t^3-3t^2+2t\,dt$

Can you proceed?

3. ## Re: Rectilinear Motion: finding displacement & distance traveled w/ integration (2nd

I will try when I get around to it today and post back but it looks good to me

I think that's what I did, I might have made some arithmetic mistake but I'll just have to try it again and post what I got if it's wrong ....

Thanks