# How do i find acceleration and x/y coordinates given time and i/j values?

piercegirl

## Homework Statement

At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of i = (3.00 i - 2.00 j) m/s and is at the origin. At t = 2.70 s, the particle's velocity is = (9.30 i + 6.90 j) m/s.

(a) Find the acceleration of the particle at any time t.
= m/s2

(b) Find its coordinates at any time t.
x = m
y = m

Homework Helper
MHB

## Homework Statement

At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of i = (3.00 i - 2.00 j) m/s and is at the origin. At t = 2.70 s, the particle's velocity is = (9.30 i + 6.90 j) m/s.

(a) Find the acceleration of the particle at any time t.
= m/s2

(b) Find its coordinates at any time t.
x = m
y = m

Welcome to PF, piercegirl!

You should have:
$$\mathbf{\vec v}_t = \mathbf{\vec v}_0 + \mathbf{\vec a} t$$
Do you?

What do you get when you fill in what you have?

piercegirl
Hello thanks for replying. Please bare with me (my professor has us doing web assign and he hasnt covered most of these topics)

→ → →
v =v(not)+at

v=(-9.81)2.70

=-26.46

This is horrible, :( Im thinking this isnt right?

Homework Helper
MHB
Hello thanks for replying. Please bare with me (my professor has us doing web assign and he hasnt covered most of these topics)

→ → →
v =v(not)+at

v=(-9.81)2.70

=-26.46

This is horrible, :( Im thinking this isnt right?

Hmm, I think you haven't got your symbols straight.
Let me list them.

##\mathbf{\vec v}_0## is the initial speed as a vector.
You have it as (3.00 i - 2.00 j).
Do you know what that "i" and "j" mean?

##\mathbf{\vec v}_t## is the speed at time t as a vector.
Do you have that?

##\mathbf{\vec a}## is the as yet unknown constant acceleration vector, which is asked for.
It is not the acceleration of gravity (which is 9.81 m/s).
Leave it as it is for now.

Can you fill in the numbers, or rather the vectors, you have in the formula?

piercegirl
I believe i is x and J is Y. How can i get the average velocity vector by x and y?

I trying to do delta r/delta t.

I was given the particles velocity 9.30x+6.90 at t=2.70s

Homework Helper
MHB
I believe i is x and J is Y. How can i get the average velocity vector by x and y?

I trying to do delta r/delta t.

I was given the particles velocity 9.30x+6.90 at t=2.70s

The i and j indeed represent x and y.
They denote a vector.
The vectors you have need to be inserted in the formula.
I'm afraid that the formula delta r/delta t won't work for your purpose.

Can you just replace the symbols by the respective vectors?

So replace v0 literally by (3.00 i - 2.00 j), etcetera, without doing anything else?

piercegirl
like this?

(9.30i+6.90j)=(3.00i-2.00j)+ a(2.70)

a=6.3i+8.9j/(2.70) i combined like terms...

Homework Helper
MHB
like this?

(9.30i+6.90j)=(3.00i-2.00j)+ a(2.70)

a=6.3i+8.9j/(2.70) i combined like terms...

Yep!
That's it.

You should include extra parentheses btw.
Like this:
a=(6.3i+8.9j)/2.70

piercegirl
yay!! :) thank you so much!!! :D now for part b. how do i find the x and y components? is it just x=6.3i and y=8.9j?

Homework Helper
MHB
yay!! :) thank you so much!!! :D now for part b. how do i find the x and y components? is it just x=6.3i and y=8.9j?

No. The x and y coordinates are not constant, so that can't be right.

You need a slightly different formula.
$$\vec x_t = \vec x_0 + \vec v_0 t + \frac 1 2 \vec a t^2$$
Did you have that formula?

Now "t" is the unknown, since you're supposed to find the result "for any t".
Can you fill in the rest?

piercegirl
ah. I did the math and i got t=2.70. Is that right?

Homework Helper
MHB
ah. I did the math and i got t=2.70. Is that right?

I haven't checked.
But that's not what they asked.
They asked: find its coordinates at any time t.
So t is not just 2.70...

piercegirl
oh! so we are solving for Vt?

Homework Helper
MHB
oh! so we are solving for Vt?

No, we are solving for Xt, that is, the position at some time t.

piercegirl
is there such a formula? It seems like the kinematic eq's keep changing :/

Homework Helper
MHB
is there such a formula? It seems like the kinematic eq's keep changing :/

See my previous post #10.
Do you have that formula?

piercegirl
see it. My apologies, for some reason i didnt see it before. I got t=.90 but idk

Homework Helper
MHB
Can you literally fill in the numbers/vectors you have in that formula without doing anything else?

piercegirl
so shall i let t= to some random num and solve for xt?

Homework Helper
MHB
Don't fill in t or xt.
Only the others, since you should have all of them.

piercegirl
sure,

xt=3i+(3.00i-2.00j)t+(1/2(6.3i+8.9j/2.70))t^2

Homework Helper
MHB
sure,

xt=3i+(3.00i-2.00j)t+(1/2(6.3i+8.9j/2.70))t^2

Almost!
Where did you get 3i?

And don't forget the parentheses around (6.3i+8.9j).
Otherwise you have something entirely different.

piercegirl
i thought x0 meant the x component of v0, being 3.00i?

Homework Helper
MHB
i thought x0 meant the x component of v0, being 3.00i?

Ah, no.
##\vec x_0## is the position at time t=0.
It is also a vector that has its own x and y coordinate.
I have to admit it's a bit confusing that x is used for the position vector as well as for an x coordinate.

Do you know what the initial position is?

piercegirl
i dont know. im thinking its (0,0)

Homework Helper
MHB
i dont know. im thinking its (0,0)

That is right!
If you read your problem statement carefully, you'll see it says: "At t = 0, (snip) and is at the origin."

x = m
y = m

Can you rewrite the formula so you'll have an answer for just the x coordinate of the position?

piercegirl
xt=(3.00i+2.00j)t+(1.16i+1.65)t2

(3.00i)t+(1.16i)t2 for x?

Homework Helper
MHB
xt=(3.00i+2.00j)t+(1.16i+1.65)t2

(3.00i)t+(1.16i)t2 for x?

Yep.
That's it!

You can leave out the "i" though.
And it's better to round 1.1666667 to 1.17.

And..... I'm off to bed! :zzz:

piercegirl
thanks!! your the best!! sweet dreams