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The acceleration of a particle on a horizontal xy plane

  • Thread starter kara
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The acceleration of a particle on a horizontal xy plane is given by a=3ti + 4tj, where a is in meters per second-squared and t is in sec. At t=0, the particle has the position vector r=(20.0m)i + (40.0m)j and the velocity vector
v=(5.00m/s)i +(2.00m/s)j. At t=4.00 s, what are its position vector in unit-vector notation and the angle between its direction of travel and the positive direction of the x axis?


I have no idea what to do at all:surprised https://www.physicsforums.com/images/smilies/surprised.gif [Broken]
:surprised
 
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2,036
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Find the general equations for position and velocity by integration and use the initial conditions.

"the angle between its direction of travel and the positive direction of the x axis"

For that you can apply the dot product.
 
54
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so would i integrate a to find position and then velocity?
 
2,036
2
You integrate accelation to find velovity, and then integrate velocity (after plugging in the initial condition) to find position.
 
54
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alright so i got:

dx = 1.5t^2 + vcos0t + 20
dy = 2t^2 + vsin0t + 40

where the 0 is theta
 
2,036
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kara said:
alright so i got:

dx = 1.5t^2 + vcos0t + 20
dy = 2t^2 + vsin0t + 40

where the 0 is theta
Bringing in new variables (such as theta) without any need will just make the answer more complicated.

Here's an example on how to go about the problem:

[tex]\frac{dv_x}{dt} = 3t[/tex]

[tex]v_x = \frac{3}{2}t^2 + C[/tex] (C is the constant of integration)

Now put in the initial condition, the value for v_x when t = 0 and find the value for C. Similaraly find v_y. The process is the same for finding position.
 
Last edited:

radou

Homework Helper
3,105
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You can be a bit more formal and use [tex]\vec{v}(t) = \int \vec{a}(t)dt + \vec{v}_{0}[/tex], i.e. [tex]\vec{r}(t) = \int \vec{v}(t)dt + \vec{r}_{0}[/tex], but basically that's the same thing Neutrino wrote.
 

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