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

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piercegirl
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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

 
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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

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?
 
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.

That is your answer for part (a).

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?
 
oh! so we are solving for Vt?
 
is there such a formula? It seems like the kinematic eq's keep changing :/
 
see it. My apologies, for some reason i didnt see it before. I got t=.90 but idk
 
so shall i let t= to some random num and solve for xt?
 
sure,

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

Now your question (b) asks for an answer of the form:
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?
 
thanks! your the best! sweet dreams
 

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