Average velocity in two dimension with given functions?

[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:

.71i + 7.371j and it isn't correct? Please help!

 

[collinsmark]

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

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.

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:

.71i + 7.371j and it isn't correct? Please help!

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.

 

[JessicaJ283782]

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