# Homework Help: Average velocity in two dimension with given functions?

1. Jan 29, 2014

### JessicaJ283782

Suppose that the position vector for a particle is given as a function of time by vector r (t) = x(t)i + y(t)j, with x(t) = at + b and y(t) = ct^2 + d, where a = 1.70 m/s, b = 1.05 m, c = 0.126 m/s2, and d = 1.12 m.

So, I evaluated x(t) and y(t) at each time:

x(t): 1.70*2.05+1.05 which equals 1.385
x(t)=1.70*3.80+1.05 which equals 1.696

y(t)=.126(2.05^2)+1.12 which equals 6.41515
y(t)=.126(3.80^2)+1.12 which equals 19.3144

I then subtracted the x coordinates (1.696-1.385)=.2975
and the y coordinated (19.3144-6.41515)=12.89925

I then divided those answers by the change in time

(.2975/1.75) and (12.88925/1.75) to get the x and y coordinates and I got:

2. Jan 29, 2014

### collinsmark

Maybe I'm missing something. Where do the times come from? I can see you are using t = 2.05 s and t = 3.80 s. but where do these come from? They're not in the problem statement.

More than that, there's something wrong with the calculations. For example,
1.70 × 2.05 + 1.05 is not equal to 1.385.

Similarly, none of the other, above calculations work out either.

That looks like the right approach if you're trying to find the average velocity from time 2.05 s to 3.80 s (as opposed to the instantaneous velocity at some particular point in time). But of course you'll first need to fix your initial calculations mentioned above.

3. Jan 29, 2014

### JessicaJ283782

Thank you very much! I'm not sure how I messed that up; thank you!