A plane flies directly from city A to city B, which are separated by 2300 mi. From A to B, the plane flies into a 65 mi/h headwind. On the return trip from B to A, the wind velocity is unchanged. The trip from B to A takes $65 \mathrm {~min}$ less than the trip from A to B. What is the airspeed (assumed constant)of the plane relative to the air (that is, the airspeed)?
A)530 mi/h
B)400 mi/h
C)480 mi/h
D)610 mi/h

Question

Quizplus · Accepted Answer

Let $v$ be the airspeed of the plane. Then the speed of the plane relative to the ground on the trip from A to B is $v-65$ mi/h, and the speed of the plane relative to the ground on the trip from B to A is $v+65$ mi/h. Since the distance from A to B is 2300 mi, the time for the trip from A to B is $\frac{2300}{v-65}$ h, and the time for the trip from B to A is $\frac{2300}{v+65}$ h. Since the trip from B to A takes 65 min less than the trip from A to B, we have$$\frac{2300}{v+65}-\frac{2300}{v-65}=\frac{65}{60}=1.083$$Solving for $v$, we get$$v=\boxed{530 \mathrm{~mi/h}}$$

A Plane Flies Directly from City a to City B 65 min65 \mathrm {~min}65 min

A Plane Flies Directly from City a to City B $65 \mathrm {~min}$