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

  • Thread starter piercegirl
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  • #1
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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

I tried following a similar thread but just jot confused. please help




Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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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

I tried following a similar thread but just jot confused. please help
Welcome to PF, piercegirl! :smile:

Let's start with the relevant equations.
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?
 
  • #3
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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?
 
  • #4
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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?
 
  • #5
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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
 
  • #6
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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?
 
  • #7
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like this?

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

a=6.3i+8.9j/(2.70) i combined like terms...
 
  • #8
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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.

That is your answer for part (a).

You should include extra parentheses btw.
Like this:
a=(6.3i+8.9j)/2.70
 
  • #9
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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?
 
  • #10
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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?
 
  • #11
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ah. I did the math and i got t=2.70. Is that right?
 
  • #12
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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...
 
  • #13
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oh! so we are solving for Vt?
 
  • #14
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oh! so we are solving for Vt?
No, we are solving for Xt, that is, the position at some time t.
 
  • #15
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is there such a formula? It seems like the kinematic eq's keep changing :/
 
  • #16
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is there such a formula? It seems like the kinematic eq's keep changing :/
See my previous post #10.
Do you have that formula?
 
  • #17
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see it. My apologies, for some reason i didnt see it before. I got t=.90 but idk
 
  • #18
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Can you literally fill in the numbers/vectors you have in that formula without doing anything else?
 
  • #19
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so shall i let t= to some random num and solve for xt?
 
  • #20
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Don't fill in t or xt.
Only the others, since you should have all of them.
 
  • #21
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sure,

xt=3i+(3.00i-2.00j)t+(1/2(6.3i+8.9j/2.70))t^2
 
  • #22
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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.
 
  • #23
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i thought x0 meant the x component of v0, being 3.00i?
 
  • #24
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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?
 
  • #25
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i dont know. im thinking its (0,0)
 

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