A roller coaster traveling at 15 m/s goes through a dip in the track whose radius of curvature is 45 m. Assuming that the acceleration of gravity is 10 m/s², the number of "g's" the riders experience is A) 1.5 g. B) 0.50 g. C) 5.0 g. D) 6.0 g.

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Quizplus · Accepted Answer

The centripetal acceleration $a_c$ experienced by the riders is given by $a_c = \frac{v^2}{r}$, where $v = 15 \, \text{m/s}$ is the velocity and $r = 45 \, \text{m}$ is the radius of curvature. Plugging in the values, we get $a_c = \frac{(15)^2}{45} = \frac{225}{45} = 5 \, \text{m/s}^2$. To find the number of "g's", we divide the centripetal acceleration by the acceleration due to gravity (10 m/s^2), resulting in $5 \, \text{m/s}^2 / 10 \, \text{m/s}^2 = 0.5 \, g$. However, this is the additional acceleration due to the curvature of the track. The total acceleration experienced by the riders is the sum of this centripetal acceleration and the acceleration due to gravity, which is $1g + 0.5g = 1.5g$.

A Roller Coaster Traveling at 15 M/s Goes Through a Dip