Solving Riemann Sum: Velocity Function v(t) = t^2 -5t + 6

[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.

 

[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).

 

[exk]

uhh the velocity function is just a simple parabola and has a single turning point on [0,5]. The difference between distance and displacement 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.

 

[HallsofIvy]

uhh the velocity function is just a simple parabola and has a single turning point on [0,5]. The difference between distance and displacement 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.

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.