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**I'm having difficult understanding the question, can somebody help me? The question is:**

"The velocity of a 3 kg particle is given by v = (3ti + 6(t^2)j )m/s, with time t in seconds. At the instant the net force on the particle has a magnitude of 35 N, what are the direction (relative to the positive direction of the x axis) of (a) the net force and (b) the particle's direction of travel?

Answer:

(a) 75.099403312170 Units ° (degrees)

(b) 61.978933315223 Units ° (degrees)"

"The velocity of a 3 kg particle is given by v = (3ti + 6(t^2)j )m/s, with time t in seconds. At the instant the net force on the particle has a magnitude of 35 N, what are the direction (relative to the positive direction of the x axis) of (a) the net force and (b) the particle's direction of travel?

Answer:

(a) 75.099403312170 Units ° (degrees)

(b) 61.978933315223 Units ° (degrees)"

My initial approach was to find t when the magnitude of the particle is 35N to get the components of the triangle.

Velocity and Acceleration

v = 3ti + 6(t^2)j

a = 3 + 12t

Time

F = ma

35 = 3(3+12t)

35 = 9 + 36t

26 = 36t

.72 = t

Since, I thought I was to find the angle created by the velocity vector, I found the y-comp. (opp.) and x-comp. (adj.).

y-comp. = 6(t^2) = 6(.72^2) = 3.13

x-comp. = 3t = 3(.72) = 2.27

tan(theta) = opp./adj. = 3.13/2.27

theta = arctan(3.13/2.27) = arctan(1.44) = 55.27

However, 55.27 isn't the angle... for anything! Did I misinterpret the question? What is the difference between the two angles they are asking me to find, what are they asking of my specifically? If somebody can help me I would deeply appreciate it.

Thank you for taking the time to read my question.