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.

not correct

To find the angle between the direction of travel and the positive x-axis, we need to determine the velocity vector of the particle at t=3.60s.

Given that the acceleration vector is a=5ti + 6tj, we can integrate this once with respect to time to find the velocity vector. So, integrating each component separately:

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

Now, we can use the initial condition at t=0 (v=(5.70m/s)i + (3.40m/s)j) to find the constants C1 and C2.

At t=0:
vx = (5/2)(0)^2 + C1 = C1
vy = 3(0)^2 + C2 = C2

Therefore, C1 = 5.70 m/s and C2 = 3.40 m/s.

Substituting these values back into the equations for vx and vy:
vx = (5/2)t^2 + 5.70
vy = 3t^2 + 3.40

Now, at t=3.60 s, we can substitute the value into these equations to find the components of the velocity vector:

vx(3.60) = (5/2)(3.60)^2 + 5.70
vy(3.60) = 3(3.60)^2 + 3.40

Calculating these values, we get:
vx(3.60) = 64.80 + 5.70 = 70.50 m/s
vy(3.60) = 3(3.60)^2 + 3.40 = 39.12 + 3.40 = 42.52 m/s

Therefore, the velocity vector at t=3.60s is v = (70.50 m/s)i + (42.52 m/s)j.

To find the angle between the direction of travel and the positive x-axis, we can calculate the arctangent of the ratio of the y-component to the x-component of the velocity vector:

θ = arctan(vy/vx)

Plugging in the values, we have:
θ = arctan((42.52 m/s) / (70.50 m/s))

Calculating this, we find:
θ ≈ 0.55 radians

Therefore, the angle between the direction of travel and the positive x-axis is approximately 0.55 radians.