How Does Particle Motion on an XY Plane Change Over Time?

[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?

 

[Locrian]

I'll say the same thing here as in the other problem: treat the axis as separate problems, solve both, then put the problem back together. By this point in class you've done 1d problems plenty - break this down into two separate 1d problems, figure out the i and j positions, then draw a picture of where the particle is and find the angle.

 

[kyle8921]

Firstly, let's break down the given information. We are given the acceleration of the particle, a=3ti + 4tj, where a is in meters per second-squared and t is in seconds. This means that the acceleration is changing over time, as it is dependent on t.

We are also given the initial position vector, r=(20.0m)i + (40.0m)j, where i and j are unit vectors in the x and y directions, respectively. This tells us that at t=0, the particle is located at the point (20.0m, 40.0m) on the xy plane.

Furthermore, we are given the initial velocity vector, v=(5.00m/s)i +(2.00m/s)j, which tells us that at t=0, the particle is moving with a velocity of 5.00 m/s in the x direction and 2.00 m/s in the y direction.

Now, we are asked to find the position vector at t=4.00 s in unit-vector notation and the angle between its direction of travel and the positive direction of the x axis. To do this, we can use the equations of motion:

r(t) = r(0) + v(0)t + 0.5at^2

v(t) = v(0) + at

Where r(t) and v(t) are the position and velocity vectors at time t, r(0) and v(0) are the initial position and velocity vectors, and a is the acceleration vector.

Substituting the given values, we have:

r(4.00 s) = (20.0m)i + (40.0m)j + (5.00m/s)i(4.00 s) + (2.00m/s)j(4.00 s) + 0.5(3ti + 4tj)(4.00 s)^2

= (20.0m + 20.0m + 24.0m)i + (40.0m + 8.00m + 16.0m)j

= (64.0m)i + (64.0m)j

Therefore, the position vector at t=4.00 s is r(4.00 s) = (64.0m)i +