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

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

Homework Equations

The Attempt at a Solution

 

[I like Serena]

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!

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?

 

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

 

[I like Serena]

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?

 

[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

 

[I like Serena]

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...

 

[I like Serena]

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

 

[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?

 

[I like Serena]

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?

 

[I like Serena]

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 V_t?

 

[I like Serena]

oh! so we are solving for V_t?

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

 

[piercegirl]

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

 

[I like Serena]

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

 

[I like Serena]

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 x_t?

 

[I like Serena]

Don't fill in t or x_t.
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

 

[I like Serena]

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?

 

[I like Serena]

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

 

[I like Serena]

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?

 

[piercegirl]

x_t=(3.00i+2.00j)t+(1.16i+1.65)t²

(3.00i)t+(1.16i)t² for x?

 

[I like Serena]

x_t=(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