# Riemman Sum Homework

1. May 3, 2008

### MillerL7

The velocity function is v(t)=t^2 -5t + 6 for a particle moving along a line. Find the displacement traveled by the particle during the time interval [0,5].

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. May 3, 2008

### DavidWhitbeck

What you did was to find the displacement, which is the integral of velocity. The distance is the integral of speed, the absolute value of velocity.

$$v(t)=(t-2)(t-3)$$ 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. May 3, 2008

### exk

uhh the velocity function is just a simple parabola and has a single turning point on [0,5]. The difference between distance and displacment in this question is that while the distance is the integral, the displacement is in fact 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. May 4, 2008

### HallsofIvy

Staff Emeritus
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.