Question Detail
Question
A 0.340 kg particle moves in an xy plane according to x(t) = -15 + 2t - 4t³ and y(t) = 25 + 7t - 9t², with x and y in metres and t in seconds. At t = 0.7 s, what are the magnitude and angle of the net force on the particle?
Correct Answer
Option B is the correct answer.
Detailed Explanation
Find acceleration components: a_x = d²x/dt² = d/dt(2 - 12t²) = -24t. At t=0.7, a_x = -24×0.7 = -16.8 m/s². a_y = d²y/dt² = d/dt(7 - 18t) = -18 m/s² (constant). F_x = m a_x = 0.34 × (-16.8) = -5.712 N. F_y = 0.34 × (-18) = -6.12 N. Magnitude F = √(5.712² + 6.12²) = √(32.63 + 37.45) = √70.08 = 8.37 N. Angle θ = tan⁻¹(F_y/F_x) = tan⁻¹(6.12/5.712) = tan⁻¹(1.071) = 47°, but both components negative, so in third quadrant, θ = -180° + 47° = -133°.
Hint
F = ma, find acceleration by differentiating twice.