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

• piercegirl
In summary: It is a vector too.It is not (3.00 i) but something else (which you still need to find).So, as you see, it is important to know what the symbols mean.okay, thank you so much! :)You're welcome!In summary, a particle starts at the origin with a velocity of (3.00 i - 2.00 j) m/s and has a velocity of (9.30 i + 6.90 j) m/s at t = 2.70 s. Using the equation v = v0 + at, the acceleration of the particle is (6.3 i + 8.9 j) m/s2. To find the coordinates at any time
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

piercegirl said:

## 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?

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, :( I am thinking this isn't right?

piercegirl said:
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, :( I am thinking this isn't 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?

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

piercegirl said:
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?

like this?

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

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

piercegirl said:
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

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?

piercegirl said:
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?

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

piercegirl said:
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...

oh! so we are solving for Vt?

piercegirl said:
oh! so we are solving for Vt?

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

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

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

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

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

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

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

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

sure,

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

piercegirl said:
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.

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

piercegirl said:
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?

i don't know. I am thinking its (0,0)

piercegirl said:
i don't know. I am 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?

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

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

piercegirl said:
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:

thanks! your the best! sweet dreams

## 1. How do I calculate acceleration using time and i/j values?

To calculate acceleration, you can use the formula a = (j - i) / t, where j and i are the initial and final velocities, and t is the time elapsed. This will give you the acceleration in units per second squared.

## 2. What are x/y coordinates and how do they relate to acceleration?

X and y coordinates refer to the position of an object on a two-dimensional plane. Acceleration is a measure of how an object's velocity changes over time, which can also affect its position on the x and y axes.

## 3. Can I find acceleration and coordinates without knowing the initial or final velocities?

No, the formula for calculating acceleration requires both the initial and final velocities. However, if you have the displacement (change in position) of the object, you can use the equation a = 2(x - x0 - vt) / t2, where x0 is the initial position, x is the final position, v is the initial velocity, and t is the time elapsed.

## 4. How can I use time and i/j values to graph an object's motion?

To graph an object's motion, you can plot the x and y coordinates at different time intervals. The slope of the resulting line will represent the object's velocity, and the slope of the line's slope (second derivative) will represent the object's acceleration.

## 5. Are there any other factors that can affect acceleration and coordinates?

Yes, there are other factors that can affect acceleration and coordinates, such as friction, air resistance, and external forces like gravity. These factors can cause the object's velocity and position to change over time, leading to a different acceleration and coordinates than what is calculated using the initial and final values.

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