The original Ferris wheel had a radius of 38 m and completed a full revolution (2$\pi$radians) every two minutes when operating at its maximum speed. If the wheel were uniformly slowed from its maximum speed to a stop in 35 seconds, what would be the magnitude of the tangential acceleration at the outer rim of the wheel during its deceleration? A)0.0015 m/s² B)0.057 m/s² C)0.54 m/s² D)1.6 m/s² E)6.8 m/s²

Question

Quizplus · Accepted Answer

The tangential acceleration can be calculated using the formula $ a_t = \frac{\Delta v}{\Delta t} $. The initial tangential velocity $ v_i $ is $ \frac{2\pi \times 38}{120} $ m/s. The final velocity $ v_f $ is 0 m/s. The time $ \Delta t $ is 35 seconds. Plugging in these values gives $ a_t = \frac{0 - \frac{2\pi \times 38}{120}}{35} \approx 0.057 $ m/s².

The Original Ferris Wheel Had a Radius of 38 M π\piπ

The Original Ferris Wheel Had a Radius of 38 M $\pi$