At t = 0, an automobile traveling north begins to make a turn. It follows one-quarter of the arc of a circle of radius 9.1 m until, at t = 1.84 s, it is traveling east. The car does not alter its speed during the turn.
To find the acceleration of the car during the turn, we can use the centripetal acceleration formula:
ac = v^2 / r
Where ac is the centripetal acceleration, v is the velocity of the car, and r is the radius of the circular path.
We are given that the car does not alter its speed during the turn. This means that the magnitude of the velocity remains constant. Therefore, to find the acceleration, we need to find the change in velocity and the time interval.
From t = 0 to t = 1.84 s, the car undergoes its turn. We know the initial velocity is in the north direction and the final velocity is in the east direction.
To find the change in velocity, we can use vector subtraction:
Δv = √((vf - vi)^2)
Where Δv is the change in velocity, vf is the final velocity, and vi is the initial velocity.
Now, we need to find the magnitudes of the initial and final velocities.
The magnitude of the initial velocity can be found using the formula:
vi = distance / time
Since the car follows one-quarter of the arc of a circle, the distance traveled can be found using the formula for the circumference of a circle:
distance = 1/4 * (2π * r)
Plugging in the radius value (9.1 m) into the formula, we get:
distance = 1/4 * (2π * 9.1)
Next, we can calculate the initial velocity:
vi = distance / time = (1/4 * (2π * 9.1)) / 1.84
Now, let's calculate the final velocity.
The magnitude of the final velocity is given by:
vf = distance / time = (1/4 * (2π * 9.1)) / 1.84
Finally, we can find the change in velocity:
Δv = √((vf - vi)^2)
Now, we have all the values needed to calculate the acceleration:
ac = Δv^2 / r
Let's plug in the values and solve for the acceleration.