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 the 5.00 s interval from the end of the fifth second to the end of the tenth second, find the particle’s displacement. Give (g) magnitude and (h) 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 (l) 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.

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 the 5.00 s interval from the end of the fifth second to the end of the tenth second, find the particle's displacement. Give (g) magnitude and (h) 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.

Answer 1

(a) At time t=5, you are at the quarter circle. It is r units to the right and r units up. Magnitude is

x = square root of (2r^2)

= square root of (2(2.40)^2)

=3.39 m

(b) Since it is r units to the right, and r units up, we can do arc tan(r/r) = 45, so 45 degrees.

(c) At this time, it is where it is shown in the picture. We know that the string forms a 45 degree. It is going by the 45-45-90 rules for right triangles, we can find out that the ball is currently r/root(2) units to the right and r + r/root(2) units up.

The magnitude is,

            x = r + r/root(2)

             = (2.40 m) +(2.40 m)/ root(2)

             =4.09 m

(d) Since we know that the ball is r/root(2) units to the right and r + r/root(2) units up, the angle is,

arc tan([r+r/root(2)]/[r/root(2)])=67.5