A particle P travels with constant speed on a circle of radius r = 2.40 m (see t

Question

A particle P travels with constant speed on a circle of radius r = 2.40 m (see the figure) and completes one revolution in 20.0 s. The particle passes through O at time t = 0. At t = 5.00 s, what is the particle's position vector? Give (a) magnitude and (b) direction (as an angle relative to the positive direction of x.) At t = 7.50 s, what is the particle's position vector? Give (c) magnitude and (d) direction (as an angle relative to the positive direction of x.) At t = 10.00 s, what is the particle's position vector? Give (e) magnitude and (f) direction (as an angle relative to the positive direction of x.)

Explanation / Answer

In 20 second 360 degree is completed. In 5 second it will complete [360/20]* 5 = 90 degree The coordinate of the point is (1.2m, 1.2m) The magnitude or the distance from the origin is v (1.2^2 + 1.2^2 ) = 1.2*v2 The angle is given by tan A = 1.2 /1.2 = 1 A = 45 degree.

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