- #1
pizzie
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- 0
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