# Distance/displacement integration problem

• zachem62
In summary, the velocity function v(t)=t^2 -2t -8, 1≤t≤6 is given and we are asked to find the displacement and distance traveled by a particle during this time interval. Integrating the velocity function from 1 to 6, we get a value of -10/3 meters. However, this only gives us the distance traveled and we still need to find the displacement. To do this, we need to consider the change in position of the particle, which can be found by subtracting the initial position from the final position. Using the given velocity function, we can find the final position when t=6 to be 16 meters and the initial position when t=1 to

## Homework Statement

Suppose that v(t)=t2 -2t -8, 1≤t≤6 is the velocity function (in meters per second) of a particle moving along a line. Find a) the displacement and b) the distance traveled by the particle during the given time interval.

## The Attempt at a Solution

so i integrated the velocity function at the interval 1 to 6 and got the answer (-10/3)m. so i have no idea if this is the distance or displacement and if it is one of those, then how would i find the other one?

thanks!

zachem62 said:

## Homework Statement

Suppose that v(t)=t2 -2t -8, 1≤t≤6 is the velocity function (in meters per second) of a particle moving along a line. Find a) the displacement and b) the distance traveled by the particle during the given time interval.

## The Attempt at a Solution

so i integrated the velocity function at the interval 1 to 6 and got the answer (-10/3)m. so i have no idea if this is the distance or displacement and if it is one of those, then how would i find the other one?

thanks!

Think about whether the particle can change its direction of travel between times 0 and 6, and if it can, think about what the integral ∫v(t): t=0..6 actually computes.

RGV

Ray Vickson said:
Think about whether the particle can change its direction of travel between times 0 and 6, and if it can, think about what the integral ∫v(t): t=0..6 actually computes.

RGV

in the question it says that t can only be between 1 and 6 so i can't use any 0. and it is possible that the particle can change direction and my guess is this means that the integral ∫v(t): t=1 to 6 computes the distance. am i right?

then how would i get displacement? can u give me a hint? i know that its the final position minus initial position but i don't know if I'm supposed to get it by integrating the function or do something else.

v(t)=t^2 -2t -8, 1≤t≤6
When t= 6, v(6)= 36- 12- 8= 16.
When t= 1, v(1)= 1- 2- 8= -11.
Now, what was the displacement? (Do you know what "displacement" means?)