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A customer sits in an amusement park ride in which the compartment is tobe pulled downward in the negative direction of a y axis with an acceleration magnitude of 1.24 g, with g=9.8 m/s². A 0.567 g coin rests on the customer's knee. Once the motion begins and in unit-vector notation,

(a) What is the coin's acceleration relative to the ground?

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Final answer:

The acceleration of the coin relative to the ground is -12.152 m/s² in the negative direction of the y-axis, which is represented as -12.152j m/s² in unit-vector notation.

Step-by-step explanation:

The student is asking about the acceleration of a coin resting on a customer's knee in an amusement park ride that is accelerating downward. The coin's acceleration relative to the ground will be the same as the compartment's acceleration if no external forces other than gravity act on the coin. In this case, the downward acceleration of the ride is given as 1.24 times the acceleration due to gravity (g).

To calculate the coin's acceleration, we multiply the acceleration due to gravity by 1.24. Since we generally take upward as positive, the acceleration of the coin relative to the ground as the compartment accelerates downward will be -1.24g.

The acceleration due to gravity is 9.8 m/s2, so the coin's acceleration in unit-vector notation, taking the downward direction as negative, is -1.24 × 9.8 m/s2 or -12.152 m/s2 in the y direction. This is represented as a = -12.152j m/s2. In the context of the amusement park ride, it is important to note that this assumes that there are no other forces acting on the coin such as wind resistance.

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