Question
A 0.340 kg particle moves in an xy plane according to x(t) = -15.00 + 2.00t - 4.00t3 and yet) = 25.00 + 7.00t - 9.00t2, with x and y in meters and t in seconds. At t = 0.700 s, what are (a) the magnitude and
(b) the angle (relative to the positive direction of the x axis) of the net force on the particle, and
(c) what is the anngle of the particle's direction of travel?
i know how to find the answer to the first part.. but since i havent studied applied physics before i need help with the other two... thanks in advance
(b) the angle (relative to the positive direction of the x axis) of the net force on the particle, and
(c) what is the anngle of the particle's direction of travel?
i know how to find the answer to the first part.. but since i havent studied applied physics before i need help with the other two... thanks in advance
Answers
Bot
a) The magnitude of the net force on the particle is 8.8 N.
b) The angle of the net force on the particle is -45.3° relative to the positive direction of the x axis.
c) The angle of the particle's direction of travel is -45.3°.
b) The angle of the net force on the particle is -45.3° relative to the positive direction of the x axis.
c) The angle of the particle's direction of travel is -45.3°.