At t = 0, an automobile traveling north begins to make a turn. It follows one-quarter of the arc of a circle of radius 10.9 m until, at t = 1.56 s, it is traveling east. The car does not alter its speed during the turn.

Find the car's speed. Find the change in its velocity during the turn. Find its average acceleration during the turn.

The distance traveled in 1/4 turn is (pi/2)*R

Divide that by the time to get the the speed.

For the change in velocity, subtract one velocity vector from the other vector at right angles. They both have the same magnitude (which is the speed).

Hint: think of the hypotenuse of an isosceles right triangle

The acceleration is V^2/R

To find the car's speed during the turn, we need to find the distance it traveled along the arc of the circle.

The length of an arc of a circle can be found using the formula s = rθ, where s is the arc length, r is the radius of the circle, and θ is the angle subtended by the arc in radians.

In this case, the radius of the circle is given as 10.9 m, and the car travels one-quarter of the arc, which corresponds to an angle of π/2 radians.

Substituting these values into the formula, we have s = (10.9 m)(π/2) = 5.45π m ≈ 17.13 m.

Since the car takes 1.56 seconds to travel this distance, its speed is given by the formula v = s/t, where v is the speed, s is the distance, and t is the time.

Substituting the values, we have v = (17.13 m)/(1.56 s) ≈ 10.99 m/s.

Therefore, the car's speed during the turn is approximately 10.99 m/s.

To find the change in velocity during the turn, we need to find the difference between the initial and final velocities of the car.

The initial velocity is the velocity when the car starts its turn, traveling north. The final velocity is the velocity at the end of the turn, when the car is traveling east.

Since the car does not alter its speed during the turn, the magnitude of the initial and final velocities is the same, but they have different directions.

To find the change in velocity, we can simply subtract the initial velocity vector from the final velocity vector.

In this case, the initial velocity vector is directed north, and the final velocity vector is directed east.

The change in velocity can be found using the Pythagorean theorem, which states that the square of the magnitude of the change in velocity is equal to the sum of the squares of the magnitudes of the initial and final velocities.

Let's denote the magnitude of the change in velocity as Δv. Then we have Δv^2 = v_i^2 + v_f^2, where v_i is the magnitude of the initial velocity and v_f is the magnitude of the final velocity.

Since the magnitudes of the initial and final velocities are the same, we can rewrite the equation as Δv^2 = 2(v_i^2).

Taking the square root of both sides, we have Δv = √(2(v_i^2)), where Δv is the magnitude of the change in velocity.

Substituting the value of the initial velocity, we have Δv = √(2(10.99 m/s)^2) ≈ 21.98 m/s.

Therefore, the change in velocity during the turn is approximately 21.98 m/s.

To find the average acceleration during the turn, we need to find the change in velocity and divide it by the time taken for the change to occur.

In this case, we have already found the change in velocity to be approximately 21.98 m/s.

The time taken for the change to occur is given as 1.56 seconds.

Therefore, the average acceleration during the turn is given by a = Δv/t, where a is the average acceleration, Δv is the change in velocity, and t is the time.

Substituting the values, we have a = (21.98 m/s)/(1.56 s) ≈ 14.11 m/s^2.

Therefore, the average acceleration during the turn is approximately 14.11 m/s^2.