# Thread: an easy particle problem

1. ## an easy particle problem

Hi, just want to double check my answers and how it is set up! My prof. grades HW very harshly :/

V(t)=t^2-2t-8 1(<=) t (<=) 6

I did this using definite integrals.

Thanks

2. Originally Posted by soyeahiknow
Hi, just want to double check my answers and how it is set up! My prof. grades HW very harshly :/

V(t)=t^2-2t-8 1(<=) t (<=) 6

I did this using definite integrals.

Thanks
Did you mean $x(t) = \int_1^6 V(t)~dt= \int_1^6 t^2-2t-8~dt$ ?

3. yep! That's what I had it to be, I worked it out and got 66.6 m

but how do I get distance traveled? would I have to take the integral of the displacement? Thanks

4. displacement = $\displaystyle \int_{t_1}^{t_2} v(t) \, dt$

distance traveled = $\displaystyle \int_{t_1}^{t_2} |v(t)| \, dt$

5. my answer for the displacement was 66.6 m

and for the distance traveled, it was 53.3 m

please let me know if I am right. I know that all the signs could have messed me up. Thanks!

6. Originally Posted by soyeahiknow
my answer for the displacement was 66.6 m

and for the distance traveled, it was 53.3 m

please let me know if I am right. I know that all the signs could have messed me up. Thanks!
in any motion problem, distance traveled $\ge$ displacement

$\displaystyle \Delta x = \int_1^6 t^2 - 2t - 8 \, dt$

$\displaystyle d = -\int_1^4 t^2 - 2t - 8 \, dt + \int_4^6 t^2 - 2t - 8 \, dt$

7. I did the problem wrong. I think I have it now

I know the difference between displacement and distance is that you have the absolute sign for distance. But in this case it doesn't matter since both integrals are positive. Am I right? Thanks

the answer for displacement is 22.3 m

and for distance, it is also 22.3 m

8. Originally Posted by soyeahiknow
I did the problem wrong. I think I have it now

I know the difference between displacement and distance is that you have the absolute sign for distance. But in this case it doesn't matter since both integrals are positive. Am I right? Thanks

the answer for displacement is 22.3 m

and for distance, it is also 22.3 m
no.

$\displaystyle \Delta x = -\frac{10}{3}$

$\displaystyle d = \frac{98}{3}$