Question

A plane flies from alphaville to betaville and then back to alphaville. when there is no wind, the round trip takes 4 hours and 48 minutes, but when there is a wind blowing from alphaville to betaville at 100 miles per hour, the trip takes 5 hours. how many miles is the distance from alphaville to betaville?

Answer 1

Refer to the diagram shown below.

d = distance (miles) from Alphaville to Betaville.
v = speed (mph) of the plane with no wind.

With no wind:
The time taken to travel a distance of 2d is 4 hrs, 48 min = 4.8 hrs.
Therefore
2d/v = 4.8
v = 2d/4.8 = 0.4167d mph (1)

With the wind:
The velocity from Alphaville to Betaville is (v + 100) mph.
The time of travel is
t₁ = d/(v+100) h
The velocity from Betaville to Alphaville is (v - 100) mph.
The time of travel is
t₂ = d/)v-100) h
Because the return trip takes 5 hours, therefore
t₁ + t₂ = 5

`(d)/(v+100) + (d)/(v-100) =5 \\ (2vd)/(v^(2)-100^(2)) =5 \\ 2vd = 5(v^(2)-10^(4))` (2)

From (1), obtain
2(0.4167)d² = 5[(0.4167d)² - 10⁴]
0.8334d² = 0.8682d² - 5 x 10⁴
0.0348d² = 5 x 10⁴
d = 1198.7 mi

Answer: 1199 miles (nearest integer)