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(a)
The average velocity of the car in the whole trip.
(a)
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Answer to Problem 47P
The average velocity of the car in the whole trip is
Explanation of Solution
Given that the radius of the circular path is
The vector diagram indicating the initial velocity
The
Write the expression for the displacement vector of the car.
Here,
The initial position vector of the car is
Write the expression for the magnitude of the displacement vector.
Here,
The average velocity will be in the same direction of the displacement vector.
Write the expression for the angle
Write the expression for the magnitude of the average velocity.
Here,
Conclusion:
Substitute
Substitute
Substitute
Substitute
Combine the magnitude and direction of the average velocity vector.
Therefore, the average velocity of the car in the whole trip is
(b)
The average acceleration of the car in the whole trip.
(b)
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Answer to Problem 47P
The average acceleration of the car in the whole trip is
Explanation of Solution
It is obtained that the average velocity of the car in the whole trip is
Write the expression for the average acceleration.
Here,
Since the car covers three quarters of the circle, the distance travelled is obtained as,
Here,
The initial velocity is directed west and the final velocity is directed south. The expression for the initial and final velocity is obtained as,
Use the expression for
Use equation (VI) in (VII).
Write the expression for the magnitude of average acceleration.
Here,
Conclusion:
From equation (VIII) it is clear that the average acceleration vector has
Substitute
Write the expression for the angle
Substitute
Substitute
Combine the magnitude and direction of the average acceleration vector.
Therefore, the average acceleration of the car in the whole trip is
(c)
The reason for which a car moving at constant speed in a circular path has a nonzero average acceleration.
(c)
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Answer to Problem 47P
The change in direction of velocity requires an acceleration, which leads to the presence of a nonzero acceleration to a car moving in a circular path with constant speed.
Explanation of Solution
The direction of instantaneous velocity of the car moving in a circular path, changes in each instant. The direction of velocity at a point on the circular path will be along the tangent drawn at each of those points. Even though the magnitude of velocity (which is the speed) of motion is constant, due to the direction change of the velocity vector, the car gains an acceleration.
Therefore, the change in direction of velocity requires an acceleration, which leads to the presence of a nonzero acceleration to a car moving in a circular path with constant speed.
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