A 0.340 kg particale moves in an xy plane according to x(t)= -15+2t-4t 3 and y(t

Question

A 0.340 kg particale moves in an xyplane according to
x(t)= -15+2t-4t3 and y(t)=25+7t-9t2, with x and y in metersand t in seconds. At t= 0.7s, what are: (a) The magnitude. (b) The angle (relative to the positive direction of thex-axis) of the net force on the particale. (c) What is the angle of the particle's direction oftravel? A 0.340 kg particale moves in an xyplane according to
x(t)= -15+2t-4t3 and y(t)=25+7t-9t2, with x and y in metersand t in seconds. At t= 0.7s, what are: (a) The magnitude. (b) The angle (relative to the positive direction of thex-axis) of the net force on the particale. (c) What is the angle of the particle's direction oftravel?

Explanation / Answer

m = 0.34 kg R = xi + yj x = -15 + 2t - 4t^3 y = 25 + 7t - 9t^2 dR/dt = vx=x' = 2 -12t^2 vy=y' = 7-18t dR'/dt = ax=x'' = -24t ay=y'' = -18 Fx = ax*m Fy = ay*m |a) Vector F | = sqrt(Fx^2 + Fy^2) |b) theta Vector F| = |Arctan(Fy/Fx)| |c) theta Vector V| = |Arctan(vy/vx)|

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