At t = 0, an automobile traveling north begins to make a turn. It follows one-quarter of the arc of a circle of radius 11.0 m until, at t = 1.86 s, it is traveling east. The car does not alter its speed during the turn.
To answer this question, we need to use the concept of angular velocity and displacement. The angular velocity (ω) of an object moving in a circular path is defined as the rate at which its angular position changes with respect to time. The angular displacement (θ) is the angle swept out by the object as it moves along the circular path.
First, let's find the angular velocity of the car during the turn. The formula to calculate angular velocity is:
ω = θ / t
where ω is the angular velocity, θ is the angular displacement, and t is the time taken to cover that angular displacement.
From the given information, we know that the car follows one-quarter of the arc of a circle, which means the angular displacement is 90 degrees or π/2 radians. The time taken to cover this displacement is 1.86 seconds.
ω = (π/2) radians / 1.86 seconds
Now, let's calculate the magnitude of the angular velocity.
ω ≈ 0.845 radians/second
The magnitude of the angular velocity is approximately 0.845 radians/second.
Since the car does not alter its speed during the turn, the linear velocity remains constant throughout the turn. The linear velocity (v) of an object moving along a circular path is related to the angular velocity (ω) using the formula:
v = r * ω
where v is the linear velocity, r is the radius of the circular path, and ω is the angular velocity.
From the given information, we know that the radius of the circular path is 11.0 meters. Therefore, we can calculate the linear velocity of the car.
v = 11.0 meters * 0.845 radians/second
Now, let's calculate the magnitude of the linear velocity.
v ≈ 9.295 meters/second
The magnitude of the linear velocity is approximately 9.295 meters/second.
So, the car's linear velocity during the turn is approximately 9.295 meters/second.