A plane flies from Penthaven to Jackson and then back to Penthaven. When there is no wind, the round trip takes $3$ hours and $20$ minutes, but when there is a wind blowing from Penthaven to Jackson at $70$ miles per hour, the trip takes $3$ hours and $50$ minutes. How many miles is the distance from Penthaven to Jackson?

(Assume that the plane flies at a constant speed, and that the turnaround time is negligible.)

gnistory Oct 12, 2024

#1**0 **

We should consider the information from both flights, put that information together, and then solve for the distance of the trip. We will take advantage of the relationship of distance, rate, and time with the formula d = rt.

Let d be the distance of the one-way trip from Penthaven to Jackson, in miles

Let r be the rate of speed of the airplane during no wind, in miles per hour

let t be the time of the one-way trip from Penthaven to Jackson, in hours

Case 1) No wind

Here, we know the total time of the round trip without wind is 3 hours and 20 minutes. Converting into hours, \(t_1 = 3 \text{ hr} + 20 \text{ min} * \frac{1 \text{ hr}}{60 \text{min}} = 3\frac{1}{3} \text{ hr} = \frac{10}{3} \text{ hr}\). Both the distance d and the rate of speed r of the trip is unknown. However, the given information concerns a round trip, which means \(2d = \frac{10}{3}r \text{ so } d = \frac{5}{3}r\).

Case 2) 70 miles per hour of wind

Here, we know the total time of the round trip with wind is 3 hours and 50 minutes. Converting into hours, \(t_2 = \text{ hr} + 50 \text{ min} * \frac{1 \text{ hr}}{60 \text{ min}} = 3\frac{5}{6} \text{ hr} = \frac{23}{6} \text{ hr}\). Both the distance d and the rate of speed r of the trip is unknown again. However, there is some relationship between rates of speeds. For example, while the plan is flying with the aide of the wind, then plane is flying at r + 70 and at r - 70 on the return flight. If d = rt, then t = d/r, so we can sum the individual times.

\(\frac{d}{r + 70} + \frac{d}{r - 70} = \frac{23}{6}\).

Now, solve for d.

\(\begin{cases} d = \frac{5}{3}r \\ \frac{d}{r + 70} + \frac{d}{r - 70} = \frac{23}{6} \end{cases} \\ r = \frac{3}{5}d \text{ so } r^2 = \frac{9}{25}d^2\\ d(r - 70)+d(r+70) = \frac{23}{6}\left(r^2 - 4900\right) \\ 6dr - 420d + 6dr + 420d = 23\left(r^2 - 4900\right) \\ 12dr = 23r^2 - 112700 \\ 12d * \frac{3}{5}d = 23*\frac{9}{25}d^2 - 112700 \\ 180d^2 = 207d^2 - 2817500\\ 27d^2 = 2817500 \\ d = \sqrt{\frac{2817500}{27}} \approx 323.035 \text{ mi}\)

The3Mathketeers Oct 13, 2024