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. For that interval, find its average velocity. Give (i) magnitude and (j) direction (as an angle relative to the positive direction of x. For that interval, find its velocity at the beginning. Give (k) magnitude and (I) direction (as an angle relative to the positive direction of x. For that interval, find its velocity at the end of the interval. Give (m) magnitude and (n) direction (as an angle relative to the positive direction of x. Find the acceleration at the beginning of that interval. Give (o) magnitude and (p) direction (as an angle relative to the positive direction of x. Next, find the acceleration at the end of that interval. Give (q) magnitude and (r) direction (as an angle relative to the positive direction of x.

Explanation / Answer

(a) since it takes 20 seconds, for one revolution, at t=5 seconds, we are the quarter circle position magnitude sqrt(2r2)=3.25

(b) r units to the right and r units up angle would be 45

(c) It is at the point P as shown in figure at 7.5 seconds, string forms a 45 degree angle with the x axis.the ball is magnitude of 3.92

(d) angle 67.5 degrees.

I have answered the first four parts as per Chegg's policy.

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