The acceleration of a particle on a horizontal xy plane

[kara]

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 https://www.physicsforums.com/images/smilies/surprised.gif

 

[neutrino]

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.

 

[kara]

so would i integrate a to find position and then velocity?

 

[neutrino]

You integrate acceleration to find velovity, and then integrate velocity (after plugging in the initial condition) to find position.

 

[kara]

alright so i got:

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

where the 0 is theta

 

[neutrino]

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:

\frac{dv_x}{dt} = 3t

v_x = \frac{3}{2}t^2 + C (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.

 

[radou]

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