Position and velocity vectors of a squirrel

[cdotter]

Homework Statement

A squirrel has x and y coordinates (1.1m, 3.4m) at time t₁=0.0s and coordinates (5.3m, -0.5m) at time t₂=3.0s. For this time interval, find (a) the components of the average velocity, and (b) the magnitude and direction of the average velocity.

Homework Equations

v_avg=\Deltax/\Deltat.

The Attempt at a Solution

X_{v avg} = (5.3m - 1.1m)/(3.0s - 0.0s) = 1.4 m/s
Y_{v avg} = (-0.5m - 3.5m)/(3.0s - 0.0s) = -1.3 m/s

I know these values are correct because I checked them in the back of the book. I don't understand how they get the magnitude and direction of the average velocity.

http://img704.imageshack.us/img704/8484/20240551.png

Here's a picture of the average velocity vectors. The resultant vector (the diagonal line) should be \sqrt{(-1.3 m/s)^2+(1.4m/s)^2}, which is 0.52 m/s. The book says it's 1.9 m/s, which means they made Y_{v avg} a positive 1.3 m/s. Why did they do this?

 

[cdotter]

Wow I'm an idiot. It would help if I did (-1.3)^2 instead of -1.3^2 like I just typed in Latex.

 

[Leptos]

I know these values are correct because I checked them in the back of the book. I don't understand how they get the magnitude and direction of the average velocity.

It's just the change in distance over the change in time(simply the time between when you start measuring the change in distance and when you stop measuring the change in distance). It's the distance between (x₁, y₁) and (x₂, y₂)

Here's a picture of the average velocity vectors. The resultant vector (the diagonal line) should be \sqrt{(-1.3 m/s)^2+(1.4m/s)^2}, which is 0.52 m/s. The book says it's 1.9 m/s, which means they made Y_{v avg} a positive 1.3 m/s. Why did they do this?

Because (-1.3)² = (-1.3)*(-1.3) = 1.69; Any number² is positive.