Force and acceleration given position equations. Find magnitude and angles

[pizzie]

Homework Statement

A 0.42 kg particle moves in an xy plane according to x(t) = - 19 + 2 t - 2 t3 and y(t) = 25 + 8 t - 7 t2, with x and y in meters and t in seconds. At t = 0.6 s, what are (a) the magnitude and (b) the angle (within (-180°, 180°] interval relative to the positive direction of the x axis) of the net force on the particle, and (c) what is the angle of the particle's direction of travel

Homework Equations

Fx = ma(x)
Fy = ma(y)

The Attempt at a Solution

I got the correct answer for part A. I found the second derivative for the x(t) and y(t)
x"(t) = -12t
y"(t) = -14

to find Fx i used the equation Fx = ma(x) = (0.42)(-12)(0.6) = -3.024
to find Fy i used Fy = ma(y) = (0.42)(-14) = -5.88

i then squared both of those values, added them together, and took the square root to get the magnitude, which is 6.61. this is the right answer.

For part B i took the inverse tangent of Fy/Fx = arctan(-5.88/-3.024) and i got 62.78 deg, but this is the wrong answer.

For C i took the inverse tangent of the first derivative equations for velocity.
Vx = 2 - 6(0.6)^2 = -0.16
Vy = 8 - 14(0.6) = -0.4
then i did arctan(-0.4/-0.16), but this is also wrong.

i'm not sure what I'm doing wrong for B and C. any help would be appreciated. thank you

 

[ehild]

Both the cosine and the sine of the angle of the force are negative. In which quadrant is that angle? The same is with the velocity. Both components are negative, what is its angle with respect to the positive x axis?

ehild

 

[pizzie]

thank you! :) :)