A particle P travels with constant speed in a circle of radius 7.3 m and completes one revolution in 34.0 s (see Figure). The particle passes through O at t = 0 s. What is the magnitude of the average velocity during the interval from t = 11.9 s and t = 16.9 s.

Question

A particle P travels with constant speed in a circle of radius 7.3 m and completes one revolution in 34.0 s (see Figure). The particle passes through O at t = 0 s. What is the magnitude of the average velocity during the interval from t = 11.9 s and t = 16.9 s.

The answer here is 1.30 m/s

I need help with:
What is the magnitude of the instantaneous velocity at t = 11.9 s?

What is the magnitude of the instantaneous acceleration at t = 16.9 s?

Thanks.

Answer 1

Constant v = 2πR/T=2π•7.3/34 = 1.3 m/s

Constant centripetal acceleration is
a=v²/R =1.3²/7.3 = 0.23 m/s²

Answer 2

To find the magnitude of the instantaneous velocity at a specific time, we can use the fact that the velocity vector of a particle moving in a circle is always tangent to the circle and perpendicular to the radius.

For the first question, to find the magnitude of the instantaneous velocity at t = 11.9 s, we need to find the distance traveled by the particle in this time interval. Since the particle completes one revolution in 34.0 s, we can find the fraction of the total distance traveled in 11.9 s:

Distance traveled in 11.9 s = (11.9 s / 34.0 s) * Circumference of the circle

The circumference of the circle is given by 2 * π * radius, so we have:

Distance traveled in 11.9 s = (11.9 s / 34.0 s) * (2 * π * 7.3 m)

Now, to find the magnitude of the instantaneous velocity, we divide the distance traveled by the time taken:

Magnitude of instantaneous velocity = Distance traveled in 11.9 s / 11.9 s

Now you can calculate these values to find the magnitude of the instantaneous velocity at t = 11.9 s.

For the second question, to find the magnitude of the instantaneous acceleration at t = 16.9 s, we can use the fact that the acceleration of a particle moving in a circle is directed toward the center of the circle and has a magnitude given by the square of the velocity divided by the radius:

Magnitude of instantaneous acceleration = (Magnitude of instantaneous velocity)^2 / radius

Using the magnitude of the instantaneous velocity found in the first question and the radius of the circle, you can calculate the magnitude of the instantaneous acceleration at t = 16.9 s.

Please note that these calculations assume that the particle's motion is uniform circular motion.