Displacement and velocity

Suppose that the velocity (in meters per second) for a particle moving along a line is given by the function v(t)=t^2-4t-12 on interval 1<t<7

a) Find the displacement

b) find the displacement traveled by the particle during the given time interval

ok for part a i know you have to take the anti-derivative and you get t^3/3 -4t^2/2 -12t and interval 1 to 7

[7^3-((4(7)^2)/2)-12(7) ] - [1^3/3-((4(1)^2)/2)-12(1)]

i got -194/3 meters

I'm not sure what to do about part b. And is my answer and method for part a correct?

Re: Displacement and velocity

In part (a), you didn't divide $\displaystyle 7^3$ by 3, so your answer should be -54 meters.

Part (b) probably should have "distance" instead of "displacement". What they want is this: the particle is moving to the left (v is negative) from t=1 to t=6. At t=6, it changes direction and moves to the right. They want the total distance traveled, which would be:

$\displaystyle \int_1^7|v(t)|\,dt=\int_1^6(-v(t))\,dt+\int_6^7v(t)\,dt$

- Hollywood

Re: Displacement and velocity

Quote:

Originally Posted by

**hollywood** In part (a), you didn't divide $\displaystyle 7^3$ by 3, so your answer should be -54 meters.

Part (b) probably should have "distance" instead of "displacement". What they want is this: the particle is moving to the left (v is negative) from t=1 to t=6. At t=6, it changes direction and moves to the right. They want the total distance traveled, which would be:

$\displaystyle \int_1^7|v(t)|\,dt=\int_1^6(-v(t))\,dt+\int_6^7v(t)\,dt$

- Hollywood

thanks! these little mathematical errors are killing me. Do we use the numbers from part a for the velocity? I really appreciate your help!

edit: i think I would do this right? :

Re: Displacement and velocity

Yes, that's correct.

- Hollywood

Re: Displacement and velocity

After plugging in everything, I got 187/3 for the integral from 1 to 6. and I got and I got 13/3 for 6 to 7 integral. Added together, gives 200/3 and that is wrong to the system. I did it multiple times and checked on two different calculators..

Re: Displacement and velocity

I get 175/3 for the integral from 1 to 6.

- Hollywood

Re: Displacement and velocity

Quote:

Originally Posted by

**hollywood** In part (a), you didn't divide $\displaystyle 7^3$ by 3, so your answer should be -54 meters.

Yes I'm annoying! (Nerd)

At some point a positive direction should be chosen. Here you are tacitly stating that the initial velocity is in the "negative" direction, or to the left side of the paper. This is what most would choose so no real problem arises here, but in a more complex problem it might lead to confusion unless you state it.

-Dan

Re: Displacement and velocity

Quote:

Originally Posted by

**hollywood** I get 175/3 for the integral from 1 to 6.

- Hollywood

thanks. For the final answer which was correct to the system I calculated to be 188/3 when you add 1 to 6 and 6 to 7

Re: Displacement and velocity

Yes, 188/3 is correct.

And yes, topsquark, part of defining our variables would be to state that positive velocity is to the right.

- Hollywood