# Homework Help: Position and velocity vectors of a squirrel

1. Jan 29, 2010

### cdotter

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
vavg=$$\Delta$$x/$$\Delta$$t.

3. The attempt at a solution
Xv avg = (5.3m - 1.1m)/(3.0s - 0.0s) = 1.4 m/s
Yv 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 [Broken]

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 Yv avg a positive 1.3 m/s. Why did they do this?

Last edited by a moderator: May 4, 2017
2. Jan 29, 2010

### 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.

3. Jan 29, 2010

### Leptos

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