- #1

- 14

- 0

What is the displacement?

What is the distance traveled?

I think that the information should look like this:

(1)(1^2 -5(1) +6)+(2)(1^2-5(2)+6)+(3)(1^2-5(3)+6)+(4)(1^2-5(4)+6)+(5)(1^2-5(5)+6) to get the distance..., but I am unsure about it.

- Thread starter MillerL7
- Start date

- #1

- 14

- 0

What is the displacement?

What is the distance traveled?

I think that the information should look like this:

(1)(1^2 -5(1) +6)+(2)(1^2-5(2)+6)+(3)(1^2-5(3)+6)+(4)(1^2-5(4)+6)+(5)(1^2-5(5)+6) to get the distance..., but I am unsure about it.

- #2

- 350

- 1

[tex]v(t)=(t-2)(t-3)[/tex] suggests that v does flip sign twice across that interval, so as long as you use a sum fine enough to have points in the interval (2,3) included in your sum, then displacement and speed should be unequal (since the velocity is negative in that interval).

- #3

- 119

- 0

if you have a negative velocity for some time and then a positive velocity for the same time you end up in the same place.

- #4

HallsofIvy

Science Advisor

Homework Helper

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The turning point is not relevant. What is relevant is that the velocity is negative between t= 2 and t= 3, positive velocity from 0 to 2 and from 3 to 5. You do NOT have "a negative velocity for some time and then a positive velocity for the same time" and the displacement is NOT 0.

if you have a negative velocity for some time and then a positive velocity for the same time you end up in the same place.

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