Question

Displacement and Distance Question: PLEASE and thank you!

Answer 1

where's the question?

Answer 2

@bahrom7893 I was trying to make it easier by attaching it..

Answer 3

Displacement is relative to starting point. Distance is just the total amount traveled.

Answer 4

Yes I know @bahrom7893 , but I don't know how to solve this in this situation..

Answer 5

Distance is the integral of absolute value of velocity, displacement is just the integral of the velocity.

Answer 6

\[\Large \text{displacement} = \int\limits\limits_{0}^{10} 8-4t dt\]
\[\Large \text{distance} = \int\limits_{0}^{10} \left| 8-4t \right|dt\]
The first is a simple integral. The second you'll have to break up into two separate integrals.

Answer 7

THANK YOU SO MUCH! This makes so much more sense @agent0smith I solved it my self but this is the best response and you're so helpful! @agent0smith

Answer 8

Haha no problem. The reason you have to break it up into two integrals is because you can't do abs value integrals by hand (without breaking them up). A calculator can do it.

Graph of the 8-4t looks like this: https://www.google.com/search?q=-4x+%2B+8&oq=-4x+%2B+8&aqs=chrome.0.69i57j69i62l3.1186j0&sourceid=chrome&ie=UTF-8

Notice it dips below the x-axis between t=0 and t=10, so you need two separate integrals for the two sections of area.

Answer 9

@skdavis93 did you solve it on your own without integration, by finding the areas under the curve...?

Answer 10

No! I used what your first reply said and then I solved it right before you posted the equations! :)

Answer 11

Oh :)