The acceleration of a particle moving only on a horizontal xy plane is given by a=5ti+6tj, where a is in meters per second-squared and t is in seconds. At t=0, the position vector r=(19.0m)i+(39.0m)j locates the particle, which then has the velocity vector v=(5.70m/s)i+(3.40m/s)j.At t=3.60s, what are (a) its position vector in unit-vector notation and (b) the angle between its direction of travel and the positive direction of the x axis?

For a) I got the answer -4.11 m/s. (This is correct)

I am having trouble finding b) though.

I apologize, I meant this for my other question. I got 22.5 for this, but it was wrong. If someone can at least show calculations in getting this, I would appreciate it.

To find the position vector of the particle at t=3.60s, we need to integrate the acceleration vector with respect to time, twice. The first integration will give us the velocity vector, and the second integration will give us the position vector.

Given acceleration vector a = 5ti + 6tj, we integrate each component separately to obtain the velocity vector v:

∫(5t)dt = (5/2)t^2 + C1
∫(6t)dt = 3t^2 + C2

The constants of integration C1 and C2 are determined by the initial conditions.
At t=0, v=(5.70m/s)i + (3.40m/s)j. So we can substitute these values to find C1 and C2:

(5/2)(0^2) + C1 = 5.70 ⟹ C1 = 5.70
3(0^2) + C2 = 3.40 ⟹ C2 = 3.40

Now we have the velocity vector:

v = (5/2)t^2 + 5.70i + 3t^2 + 3.40j
v = (5/2)t^2 + 3t^2 + 5.70i + 3.40j
v = (13/2)t^2 + 5.70i + 3.40j

Next, we integrate each component of the velocity vector to find the position vector r:

∫[(13/2)t^2]dt = (13/6)t^3 + C3
∫(5.70)dt = 5.70t + C4
∫(3.40)dt = 3.40t + C5

Again, the constants of integration C3, C4, and C5 are determined by the initial conditions.
At t=0, r = (19.0m)i + (39.0m)j. So we can substitute these values to find C3, C4, and C5:

(13/6)(0^3) + C3 = 19.0 ⟹ C3 = 19.0
5.70(0) + C4 = 39.0 ⟹ C4 = 39.0
3.40(0) + C5 = 0 ⟹ C5 = 0

Now we have the position vector:

r = (13/6)t^3 + 19.0i + 5.70t + 39.0j

To find the position vector at t=3.60s, we substitute t=3.60 into the position vector:

r = (13/6)(3.60^3) + 19.0i + 5.70(3.60) + 39.0j
r = (13/6)(46.656) + 19.0i + 20.52 + 39.0j
r = (78.48) + 19.0i + 59.52 + 39.0j
r = 19.0i + 39.0j + 19.0i + 39.0j + 78.48
r = 38.0i + 78.0j + 78.48

So, the position vector at t = 3.60s is (38.0m)i + (78.0m)j + (78.48m).

To find the angle between its direction of travel and the positive x-axis, we can consider the velocity vector at t=3.60s, which is the derivative of the position vector with respect to time.

v = (13/2)t^2 + 5.70i + 3.40j

At t = 3.60s, we substitute t=3.60 into the velocity vector:

v = (13/2)(3.60^2) + 5.70i + 3.40j
v = (13/2)(46.656) + 5.70i + 3.40j
v = (78.48) + 5.70i + 3.40j
v = 5.70i + 3.40j + 78.48

Now, we can find the angle between this vector and the positive x-axis using the arctan function:

θ = arctan(vy/vx)
θ = arctan(3.40/5.70)
θ ≈ 30.49 degrees

Therefore, the angle between the direction of travel and the positive x-axis is approximately 30.49 degrees.