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Finding coordinates after finding the acceleration of an object

  • #1
Acceleration and coordinates at time t.

Homework Statement


At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of vector v i = (3.00 i - 2.00 j) m/s and is at the origin. At t = 3.60 s, the particle's velocity is vector v = (7.00 i + 3.70 j) m/s. (Use the following as necessary: t. Round your coefficients to two decimal places.)

a) Find the acceleration of the particle at any time t. Found this a= (4.00i+5.70j)/3.6s
b) Find its coordinates at any time t. I have no clue how to do this

Homework Equations



I tried x⃗t=x⃗0+v⃗0t+12a⃗t^2 but webassign says my answer is wrong

The Attempt at a Solution


So i used the above equation and plugged in what I am given or know.
Xo=0
Vot=(3.00i-2.00j)t
a=(4.00i+5.70j)/3.60
soooo
Xt= 0 + (3.00i-2.00j)t + 1/2[(4.00i+5.70j)/3.60]t^2
in the end i got
x=(3.00i)t + (0.56i)t^2
y=(2.00j)t + (0.79j)t^2

Can someone help me understand why i got this wrong?
 
Last edited:

Answers and Replies

  • #2
32
1
a = 1.11 i + 1.58 j
Integrating,
v = 1.11 t i + 1.58 t j
Again, integrating,
s = 1.11 t[itex]^{2}[/itex]/2 i + 1.58 t[itex]^{2}[/itex]/2 j
 
  • #3
a = 1.11 i + 1.58 j
Integrating,
v = 1.11 t i + 1.58 t j
Again, integrating,
s = 1.11 t[itex]^{2}[/itex]/2 i + 1.58 t[itex]^{2}[/itex]/2 j
I understand where the a= 1.11i + 1.58j comes from but I dont understand what the other two parts (v and s) have to do with this
Can you explain?
 
  • #4
32
1
s = 1.11 t[itex]^{2}[/itex]/2 i + 1.58 t[itex]^{2}[/itex]/2 j
so
x = 1.11 t[itex]^{2}[/itex]/2
y = 1.58 t[itex]^{2}[/itex]/2
 
  • #5
s = 1.11 t[itex]^{2}[/itex]/2 i + 1.58 t[itex]^{2}[/itex]/2 j
so
x = 1.11 t[itex]^{2}[/itex]/2
y = 1.58 t[itex]^{2}[/itex]/2
So your saying I have no need to use Xt= Xo + Vot + 1/2at^2?
But what is s?
Cause it looks like your saying that by using the i and j components of "s", the x and y would be constants
but thats not possible is it? because im looking for the coordinates at any time t
 
  • #6
32
1
Unfortunately I never used proper limits in integration. Anyway lets do it with your equation.
By s I meant displacement vector.
Now,
[itex]\overline{s}[/itex] = [itex]\overline{v0}[/itex]t + 1/2 [itex]\overline{a}[/itex]t[itex]^{2}[/itex]
[itex]\Rightarrow[/itex] [itex]\overline{s}[/itex] = (3i - 2j)t + 1/2 (1.11i + 1.58j)t[itex]^{2}[/itex]
[itex]\Rightarrow[/itex] [itex]\overline{s}[/itex] = (3t +.56t^2)i + (-2t + .79t^2)j
Thus finally,
x = (3t + .56t^2)
y = (-2t + .79t^2)
 
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  • #7
ohhh ok that makes sense
I almost had it right but you cleared up my confusion
thank you so much!!
 
  • #8
32
1
Correcting my first approach:

[itex]\bar{a}[/itex] = 1.11i + 1.58j
Integrating, [itex]\bar{v}[/itex] - [itex]\bar{v0}[/itex] = 1.11ti + 1.58tj
Again integrating, [itex]\bar{s}[/itex] - [itex]\bar{s0}[/itex]- [itex]\bar{v0}[/itex]t = 1.11 (t^2)/2 i + 1.58 (t^2)/2 j
But [itex]\bar{s0}[/itex] = 0 as the particle is at origin at t = 0
So, [itex]\bar{s}[/itex] = [itex]\bar{v0}[/itex]t + 1.11 (t^2)/2 i + 1.58 (t^2)/2 j
...
 
Last edited:
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