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While riding a merry-go-round, a passenger is traveling at a speed of 1.3 m/s and is experiencing an acceleration of 1 m/s2. Later in the ride, the merry-go-round has increased its speed so that the passenger is traveling at 2.6 m/s. What acceleration does this same passenger experience now?

Respuesta :

skyluke89
The passenger in the merry-go-round is moving by uniform circular motion, and its centripetal acceleration is given by:
[tex]a= \frac{v^2}{r} [/tex]
where v is the speed of the passenger and r the radius of the merry-go-round.

We can calculate the radius of the trajectory using the velocity and the acceleration of the first part of the problem:
[tex]r= \frac{v^2}{a}= \frac{(1.3 m/s)^2}{1 m/s^2}=1.69 m [/tex]

And since the radius of the trajectory does not change, we can use it now to calculate the centripetal acceleration of the passenger when its speed has increased to 2.6 m/s:
[tex]a= \frac{v^2}{r}= \frac{(2.6 m/s)^2}{1.69 m}=4 m/s^2 [/tex]
iriantojasmine

Answer:

Correct Answer: D. 4 m/s2

Explanation:

Explanation from Gizmo: The acceleration of an object in uniform circular motion is proportional to the square of the speed at which the object travels. This means that, if the speed of an object is multiplied by a number x, the acceleration of that object is multiplied by x2. If the speed of the object is doubled, as it was in this case (from 1.3 m/s to 2.6 m/s), the acceleration will be multiplied by 4, from 1 m/s2 to 4 m/s2.

Got it correct on Uniform Circular Motion Gizmo : ExploreLearning Study

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