At \( 9: 13 \) A.M., a sports car is traveling 34 miles per hour. Two minutes later, the car is traveling 90 miles per hour The Mean Value theorem guarantees that at some time during this two-minute interval, the car's acceleration is exactly A miles per hour squared. Compute the value of \( A \). \( A=\square \mathrm{mi} / \mathrm{h}^{2} \) Need Help? Readit ASK YOUR TEACHER

To find the acceleration \( A \), we can use the Mean Value Theorem, which states that if a function is continuous over a closed interval and differentiable over the open interval, there exists at least one point where the instantaneous rate of change (the derivative) equals the average rate of change over that interval. First, we calculate the change in speed over the time interval of 2 minutes (which is \( \frac{2}{60} \) hours or \( \frac{1}{30} \) hours). Initial speed: \( 34 \) mph Final speed: \( 90 \) mph Change in speed: \( 90 - 34 = 56 \) mph Now, we find the average acceleration over this interval: \[ \text{Average acceleration} = \frac{\text{Change in speed}}{\text{Change in time}} = \frac{56 \text{ mph}}{\frac{1}{30} \text{ hours}} = 56 \text{ mph} \times 30 = 1680 \text{ mi/h}^2 \] Thus, the value of \( A \) is \( 1680 \text{ mi/h}^2 \). A = 1680 mi/h²