A particle P travels with constant speed in a circle of radius 8.6 m and completes one revolution in 12.0 s. The particle passes through O at t = 0 s. What is the magnitude of the average velocity during the interval from t = 5.4 s and t = 9.7 s?
What is the magnitude of the instantaneous velocity at t = 5.4 s?
What is the magnitude of the instantaneous acceleration at t = 9.7 s?
I have attempted this problem several times and cannot seem to arrive at the correct answers, so if someone could give me the answers, I would greatly appreciate it, and I could work backwards using my formulas. Thanks
Between 5.4 and 9.7s, the particle completes 4.3/12.0 = 35.8% of a revolution, or 129 degrees of travel around the circle. Displacement during that interval is 2R*sin(129/2)= 15.52 m
Average velocity during the interval is that displacement divided by 4.3s.
For instantaneous velocity, multiply R by the angular speed w = 2 pi/12 = 0.5236 rad/s
Instantaneous velocity magnitude is
a = R w^2.
It is centripetal only. There is no tangential acceleration along the circle
To find the answers to these questions, we can use the following formulas:
1. The average velocity is given by the displacement divided by the time taken.
2. The magnitude of the instantaneous velocity is given by the magnitude of the particle's velocity vector at a specific time.
3. The magnitude of the instantaneous acceleration is given by the magnitude of the particle's acceleration vector at a specific time.
Now, let's calculate the answers step by step:
1. To find the magnitude of the average velocity during the interval from t = 5.4 s and t = 9.7 s, we need to find the displacement first.
The particle completes one revolution in 12.0 s, so in 9.7 s, it completes (9.7 s / 12.0 s) = 0.8083 revolutions.
The circumference of the circle is 2πr, where r is the radius. Using the given radius of 8.6 m, the circumference is 2π(8.6) ≈ 54.06 m.
The displacement is the distance traveled along the circumference in 0.8083 revolutions, which is (0.8083)(54.06 m) = 43.74 m.
The time taken is the difference between the final and initial times, which is (9.7 s - 5.4 s) = 4.3 s.
Therefore, the magnitude of the average velocity is (43.74 m / 4.3 s) ≈ 10.16 m/s.
So, the magnitude of the average velocity during the interval from t = 5.4 s and t = 9.7 s is approximately 10.16 m/s.
2. To find the magnitude of the instantaneous velocity at t = 5.4 s, we will calculate the magnitude of the particle's velocity vector.
The particle completes one revolution in 12.0 s. Therefore, in 5.4 s, it completes (5.4 s / 12.0 s) = 0.45 revolutions.
The magnitude of the velocity vector is given by v = (2πr)(1 / T), where T is the period (time for one revolution).
Substituting the given values, we have v = (2π(8.6)) / 12.0 ≈ 4.524 m/s.
Therefore, the magnitude of the instantaneous velocity at t = 5.4 s is approximately 4.524 m/s.
3. To find the magnitude of the instantaneous acceleration at t = 9.7 s, we need to consider the change in velocity over time.
Since the particle is traveling at a constant speed, the magnitude of the acceleration is zero.
Therefore, the magnitude of the instantaneous acceleration at t = 9.7 s is 0 m/s².
In summary:
- The magnitude of the average velocity during the interval from t = 5.4 s and t = 9.7 s is approximately 10.16 m/s.
- The magnitude of the instantaneous velocity at t = 5.4 s is approximately 4.524 m/s.
- The magnitude of the instantaneous acceleration at t = 9.7 s is 0 m/s².
To find the magnitude of the average velocity during the interval from t = 5.4 s to t = 9.7 s, you can use the formula:
Average velocity = Total displacement / Total time
In this case, the particle travels in a circle, so the displacement is equal to the circumference of the circle. The circumference of a circle is given by 2πr, where r is the radius.
So, the displacement is equal to 2π(8.6) = 54.02 m, and the total time is 9.7 s - 5.4 s = 4.3 s.
Therefore, the average velocity is 54.02 m / 4.3 s = 12.56 m/s.
To find the magnitude of the instantaneous velocity at t = 5.4 s, you can calculate the displacement from O to the particle at that instant. Since the particle completes one full revolution in 12.0 s, at t = 5.4 s, it has completed 5.4 s / 12.0 s = 0.45 of a revolution.
Since there are 2π radians in one revolution, the particle has traveled a distance of 0.45 * 2π * 8.6 m = 12.167 m.
Therefore, the magnitude of the instantaneous velocity at t = 5.4 s is equal to the magnitude of the displacement divided by the time interval: 12.167 m / 5.4 s = 2.26 m/s.
To find the magnitude of the instantaneous acceleration at t = 9.7 s, you can use the fact that acceleration is always directed towards the center of the circle and has magnitude equal to the square of the velocity divided by the radius:
Instantaneous acceleration = (Instantaneous velocity)^2 / Radius
Using the instantaneous velocity at t = 9.7 s (which we need to calculate), we can determine the magnitude of the instantaneous acceleration.
To calculate the instantaneous velocity at t = 9.7 s, we can use the fact that the particle completes one full revolution in 12.0 s. Therefore, at t = 9.7 s, the particle has completed 9.7 s / 12.0 s = 0.8083 of a revolution.
This means the particle has traveled a distance of 0.8083 * 2π * 8.6 m = 17.203 m.
Therefore, the magnitude of the instantaneous velocity at t = 9.7 s is equal to the magnitude of the displacement divided by the time interval: 17.203 m / 9.7 s = 1.776 m/s.
Now, we can calculate the magnitude of the instantaneous acceleration using this velocity:
Instantaneous acceleration = (1.776 m/s)^2 / 8.6 m = 0.367 m/s^2.
So, the magnitude of the instantaneous acceleration at t = 9.7 s is 0.367 m/s^2.
I hope this explanation helps you understand how to solve the problem!