Name102's question at Yahoo Answers regarding displacement vs. distance

[MarkFL]

Here is the question:

The velocity function (in meters per second) for a particle moving along a line is v(t)=2t−8?

Over the time interval 0≤t≤10.

1)Find the displacement of the particle over the time interval.

2) Find the total distance traveled by the particle over the time interval.

Here is a link to the question:

The velocity function (in meters per second) for a particle moving along a line is v(t)=2t−8? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello Name102,

First, let's look at a graphical method, involving simple geometry, which we can use to answer the questions. Let's graph the linear velocity function, and take the magnitude of the area between the $x$-axis and the velocity function as distance, and applying the sign for displacement.

The area below is shaded in red and the area above is shaded in green:

Now to find the displacement we subtract the red area from the green and for the distance we add the two areas. This is easy to find since they are both right triangles.

The red area is:

\(\displaystyle A_R=\frac{1}{2}\cdot4\cdot8=16\)

The green area is:

\(\displaystyle A_G=\frac{1}{2}\cdot6\cdot12=36\)

1.) The displacement of the particle is then \(\displaystyle A_G-A_R=36-16=20\)

2.) The total distance traveled is then \(\displaystyle A_G+A_R=36+16=52\)

However, only in cases of constant acceleration will it be this simple, so let's examine how we may do this by applying the calculus.

We may orient our coordinate axis such that the initial position of the particle coincides with the origin.

For displacement, we have the IVP:

\(\displaystyle \frac{dx}{dt}=2t-8\) where \(\displaystyle x(0)=0\)

Integrating with respect to $t$, we obtain:

\(\displaystyle \int\,dx=\int 2t-8\,dt\)

\(\displaystyle x(t)=t^2-8t+C\)

Using the initial values, we may determine the value of the parameter $C$:

\(\displaystyle x(0)=0^2-8(0)+C=0\,\therefore\,C=0\)

and so the solution satisfying the IVP is:

\(\displaystyle x(t)=t^2-8t\)

Now, the displacement is the final position minus the initial position:

\(\displaystyle x(10)-x(0)=20-0=20\)

For total distance, we have the IVP:

\(\displaystyle \frac{dx}{dt}=|2t-8|\) where \(\displaystyle x(0)=0\)

Here, we may treat the function as piecewise and use two integrals:

\(\displaystyle \int\,dx=\int_0^4 8-2t\,dt+\int_4^{10}2t-8\,dt\)

\(\displaystyle x=\left[8t-t^2 \right]_0^4+\left[t^2-8t \right]_4^{10}=(16-0)+(20-(-16))=52\)

To Name102 and any other guests viewing this topic, I invite and encourage you to post other calculus questions here in our http://www.mathhelpboards.com/f10/ forum.

Best Regards,

Mark.