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Chapter 17: Problem 20

A roller-coaster reaches a velocity of \(20 \mathrm{~m} / \mathrm{s}\) at a location where the radius of curvature is \(40 \mathrm{~m}\). Calculate the acceleration, in m/s \({ }^{2}\). a) 8 c) 10 b) 9 d) 12

Short Answer

Expert verified
Answer: The acceleration of the roller-coaster at this location is 10 m/s².

Step by step solution

01

Write the centripetal acceleration formula

The formula for centripetal acceleration is given by: $$a_c = \dfrac{v^2}{r}$$ where \(a_c\) is the centripetal acceleration, \(v\) is the velocity of the object, and \(r\) is the radius of curvature.
02

Substitute the given values

Now, substitute the given values of velocity (\(v = 20 \mathrm{~m} / \mathrm{s}\)) and radius of curvature (\(r = 40 \mathrm{~m}\)) into the formula. $$a_c = \dfrac{(20\mathrm{~m} / \mathrm{s})^2}{40\mathrm{~m}}$$
03

Calculate the acceleration

Now, simplify and calculate the centripetal acceleration: $$a_c = \dfrac{400 \mathrm{~m}^2 / \mathrm{s}^2}{40\mathrm{~m}} = 10 \mathrm{~m} / \mathrm{s}^2$$ The acceleration of the roller-coaster at this location is \(10 \mathrm{~m} / \mathrm{s}^2\). So, the correct answer is (c) 10.

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