# an easy particle problem

• Jul 13th 2010, 07:49 PM
soyeahiknow
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
• Jul 13th 2010, 08:06 PM
pickslides
Quote:

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

• Jul 14th 2010, 03:59 PM
soyeahiknow
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
• Jul 14th 2010, 04:09 PM
skeeter
displacement = $\displaystyle \int_{t_1}^{t_2} v(t) \, dt$

distance traveled = $\displaystyle \int_{t_1}^{t_2} |v(t)| \, dt$
• Jul 18th 2010, 11:57 AM
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!
• Jul 18th 2010, 12:49 PM
skeeter
Quote:

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

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

$\displaystyle d = -\int_1^4 t^2 - 2t - 8 \, dt + \int_4^6 t^2 - 2t - 8 \, dt$
• Jul 18th 2010, 01:11 PM
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
• Jul 18th 2010, 01:29 PM
skeeter
Quote:

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.

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

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