Motion in a Circle
A particle P travels with constant speed in a circle of radius 5.6 m and completes one revolution in 36.0 s. The particle passes through O at t = 0 s.
What is the magnitude of the average velocity during the interval from t = 3.6 s and t = 12.5 s.
What is the magnitude of the instantaneous velocity at t = 3.6 s?
What is the magnitude of the instantaneous acceleration at t = 12.5 s?
To find the magnitude of the average velocity during the interval from t = 3.6 s and t = 12.5 s, we can use the formula:
Average velocity = displacement / time interval
The displacement of the particle during this interval can be calculated as the difference between the position vector at t = 12.5 s and t = 3.6 s. Let's call the position vector at t = 3.6 s as r_initial and the position vector at t = 12.5 s as r_final.
r_initial = 5.6 m (cos(2π(3.6/36) + i sin(2π(3.6/36))
r_final = 5.6 m (cos(2π(12.5/36) + i sin(2π(12.5/36))
Let's find the values of r_initial and r_final:
r_initial = 5.6 m (cos(0.2π) + i sin(0.2π)) ≈ 4.02 + 4.79i m
r_final = 5.6 m (cos(0.694π) + i sin(0.694π)) ≈ -2.67 - 5.43i m
Now, let's calculate the displacement:
Displacement = r_final - r_initial
= (-2.67 - 5.43i) - (4.02 + 4.79i)
= -2.67 - 5.43i - 4.02 - 4.79i
= -6.69 - 10.22i
The magnitude of the displacement is given by:
|Displacement| = √((-6.69)^2 + (-10.22)^2)
≈ √(44.5761 + 104.6884)
≈ √149.2645
≈ 12.21 m
The time interval is 12.5 s - 3.6 s = 8.9 s
Therefore, the magnitude of the average velocity during the interval from t = 3.6 s and t = 12.5 s is:
Average velocity = |Displacement| / time interval
= 12.21 m / 8.9 s
≈ 1.37 m/s
Now, let's find the magnitude of the instantaneous velocity at t = 3.6 s:
Instantaneous velocity is given by the derivative of the position vector with respect to time.
v(t) = dr(t) / dt
Let's compute the derivative of the position vector:
v(t) = 5.6 m [d(cos(2π(t/36)) + i sin(2π(t/36))) / dt]
= 5.6 m [(-2π/36)(sin(2π(t/36))) + i(2π/36)(cos(2π(t/36)))]
Plugging in t = 3.6 s into the expression, we have:
v(3.6) = 5.6 m [(-2π/36)(sin(2π(3.6/36))) + i(2π/36)(cos(2π(3.6/36)))]
= 5.6 m [(-2π/36)(sin(2π/36)) + i(2π/36)(cos(2π/36))]
Now, calculating the values:
v(3.6) ≈ -0.5519 + 0.1659i m/s
The magnitude of the instantaneous velocity at t = 3.6 s is given by:
|v(3.6)| = √((-0.5519)^2 + (0.1659)^2)
≈ √(0.30417061 + 0.02753781)
≈ √0.33170842
≈ 0.576 m/s
Moving on to finding the magnitude of the instantaneous acceleration at t = 12.5 s:
Instantaneous acceleration is given by the derivative of the instantaneous velocity with respect to time.
a(t) = dv(t) / dt
Let's compute the derivative of the instantaneous velocity:
a(t) = d(5.6 m [(-2π/36)(sin(2π(t/36))) + i(2π/36)(cos(2π(t/36)))]) / dt
= 5.6 m [-(2π/36)^2)(cos(2π(t/36))) + i(-(2π/36)^2)(sin(2π(t/36)))]
We can plug in t = 12.5 s into the expression, which gives:
a(12.5) = 5.6 m [-(2π/36)^2)(cos(2π(12.5/36))) + i(-(2π/36)^2)(sin(2π(12.5/36)))]
Calculating the values:
a(12.5) ≈ -0.0024 - 0.0139i m/s^2
The magnitude of the instantaneous acceleration at t = 12.5 s is given by:
|a(12.5)| = √((-0.0024)^2 + (-0.0139)^2)
≈ √(0.00000576 + 0.00019321)
≈ √0.00019897
≈ 0.0141 m/s^2
To find the magnitude of the average velocity during the interval from t = 3.6 s and t = 12.5 s, we can use the formula:
Average velocity = Total displacement / Total time
In this case, the total displacement is equal to the distance traveled along the circular path. Since the particle completes one full revolution in 36.0 s, the distance traveled is equal to the circumference of the circle, which is given by:
Circumference = 2 * π * radius
Plugging in the values, we get:
Circumference = 2 * π * 5.6 m = 2 * 3.14 * 5.6 m ≈ 35.2 m
Next, we need to find the total time taken to travel from t = 3.6 s to t = 12.5 s, which is equal to the difference between the two timestamps:
Total time = 12.5 s - 3.6 s = 8.9 s
Now we can calculate the average velocity:
Average velocity = (Total displacement) / (Total time) = 35.2 m / 8.9 s ≈ 3.96 m/s
Therefore, the magnitude of the average velocity during the interval from t = 3.6 s and t = 12.5 s is approximately 3.96 m/s.
To find the magnitude of the instantaneous velocity at t = 3.6 s, we can use the fact that the particle travels with constant speed along the circular path. The magnitude of the instantaneous velocity is equal to the magnitude of the constant speed, which is equal to the distance traveled in one revolution divided by the time taken to complete one revolution:
Instantaneous velocity = (Distance traveled in one revolution) / (Time taken for one revolution)
We already found the distance traveled in one revolution, which is the circumference of the circle (35.2 m), and the time taken for one revolution is given as 36.0 s. Plugging in the values, we get:
Instantaneous velocity = 35.2 m / 36.0 s ≈ 0.978 m/s
Therefore, the magnitude of the instantaneous velocity at t = 3.6 s is approximately 0.978 m/s.
To find the magnitude of the instantaneous acceleration at t = 12.5 s, we can use the formula for centripetal acceleration:
Instantaneous acceleration = (Instantaneous velocity)² / radius
We already found the instantaneous velocity at t = 12.5 s, which is the magnitude of the constant speed (0.978 m/s). Plugging in the values, we get:
Instantaneous acceleration = (0.978 m/s)² / 5.6 m ≈ 0.169 m/s²
Therefore, the magnitude of the instantaneous acceleration at t = 12.5 s is approximately 0.169 m/s².