Suppose a car has a 100 000-mile 10-year warranty on its wheels and axles.How may turns will the wheels of 0.70-m diameter make if both the mileage and time warranties run out simultaneously and what will be the average angular velocities of the wheel? Assume that a mile = 1.6 × 10³ m and that a year is 3.16 × 10⁷s.

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Quizplus · Accepted Answer

First, calculate the total distance traveled in meters: $100,000 \text{ miles} \times 1.6 \times 10^6 \text{ m/mile} = 1.6 \times 10^{8} \text{ m}$. The circumference of a wheel is $\pi d = \pi \times 0.7 \text{ m}$. The number of turns is the total distance divided by the circumference: $1.6 \times 10^{8} \text{ m} / (\pi \times 0.7 \text{ m}) \approx 7.3 \times 10^{7}$ turns. For the average angular velocity, $\omega = \Delta \theta / \Delta t$, where $\Delta \theta = 2\pi \times 7.3 \times 10^{7}$ (since each turn is $2\pi$ radians) and $\Delta t = 10 \text{ years} = 10 \times 3.16 \times 10^{7} \text{ s}$. Thus, $\omega \approx 1.4 \text{ rad/s}$.

Suppose a Car Has a 100 000-Mile 10-Year Warranty on Its