A wheel, starting from rest, has a constant angular acceleration of 1.8 rad/s2. In a 3.7-s interval, it turns through an angle of 52 rad. How long has the wheel been in motion at the start of this 3.7-s interval?

The wheel has been in motion for 2.9 s at the start of the 3.7-s interval. This can be calculated using the equation for angular displacement: θ = ω0t + 1/2αt2, where θ is the angular displacement, ω0 is the initial angular velocity (which is 0 in this case), t is the time, and α is the angular acceleration. Rearranging the equation to solve for t gives t = 2θ/α. Plugging in the given values gives t = 2(52 rad)/(1.8 rad/s2) = 2.9 s.

To find the time the wheel has been in motion at the start of the 3.7-second interval, we can use the equation:

θ = ω₀t + (1/2)αt²

where:
θ = angle turned (52 rad)
ω₀ = initial angular velocity (0 rad/s, since the wheel starts from rest)
α = angular acceleration (1.8 rad/s²)
t = time

Rearranging the equation to solve for t, we get:

0.5αt² + ω₀t - θ = 0

Substituting the given values, the equation becomes:

0.5(1.8)t² + 0t - 52 = 0

Now, we can solve this quadratic equation using the quadratic formula:

t = (-b ± √(b² - 4ac)) / (2a)

where a = 0.5α, b = ω₀, and c = -θ.

Plugging in the values, we have:

t = (0 - √(0² - 4(0.5(1.8)(-52)))) / (2(0.5(1.8)))
t = (- √(0 + 4(0.5)(1.8)(52))) / (2(0.9))
t = (- √(0 + 4(0.9)(52))) / (2(0.9))
t = (- √(0 + 187.2)) / (1.8)
t = (- √(187.2)) / (1.8)
t ≈ -8.6 s or t ≈ 8.6 s

Since time cannot be negative, the wheel has been in motion for approximately 8.6 seconds at the start of the 3.7-second interval.

To find the time the wheel has been in motion at the start of the 3.7-second interval, we can use kinematic equations of rotational motion.

First, we need to find the initial angular velocity (ω0) of the wheel. Since the wheel starts from rest, its initial angular velocity is 0 rad/s.

Next, we can use the following equation to find the final angular velocity (ω) of the wheel after a given time:

ω = ω0 + αt

Where ω is the final angular velocity, ω0 is the initial angular velocity, α is the angular acceleration, and t is the time interval.

In this case, we know that the angular acceleration (α) is 1.8 rad/s^2, and the time interval (t) is 3.7 seconds. We can substitute these values into the equation:

ω = 0 + (1.8 rad/s^2)(3.7 s)
ω ≈ 6.66 rad/s

Now, we can use another kinematic equation to find the angular displacement (θ) of the wheel during this 3.7-second interval:

θ = ω0t + (1/2)αt^2

Since the wheel starts from rest (ω0 = 0), the equation simplifies to:

θ = (1/2)αt^2

Substituting the known values:

52 rad = (1/2)(1.8 rad/s^2)(3.7 s)^2

Now, we can solve for the unknown, which is the time the wheel has been in motion at the start of the 3.7-second interval (t0).

Rearranging the equation:

t0^2 = (2 * 52 rad) / (1.8 rad/s^2)
t0^2 ≈ 57.78 s^2

Taking the square root of both sides:

t0 ≈ sqrt(57.78 s^2)
t0 ≈ 7.61 s

Therefore, the wheel has been in motion for approximately 7.61 seconds at the start of this 3.7-second interval.