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An executive chartered a private aircraft to complete a trip between two cities separated by a non-stop distance of 600600 miles. On the return flight along the exact same route, strong headwinds reduced the aircraft's average ground speed by 5050 miles per hour compared to its outbound speed. As a result, the return flight took 22 hours longer than the outbound flight. What was the average speed, in miles per hour, of the aircraft on the outbound flight?

Cevap: 150 miles per hour

Cevap

150 miles per hour
By setting the outbound speed to vv and return speed to v−50v - 50, the relationship between outbound flight time 600v\frac{600}{v} and return flight time 600v−50\frac{600}{v - 50} yields the rational equation 600v−50−600v=2\frac{600}{v - 50} - \frac{600}{v} = 2. Solving this equation gives the quadratic v2−50v−15,000=0v^2 - 50v - 15,000 = 0, which factors as (v−150)(v+100)=0(v - 150)(v + 100) = 0. The positive root gives an outbound speed of 150 miles per hour.

Adım Adım Çözüm

1
Define variables for the unknown outbound rate and express travel times for both legs.
Let vv be the outbound speed in miles per hour. Outbound time is 600v\frac{600}{v} hours and return time is 600v−50\frac{600}{v - 50} hours.
Distance equals speed multiplied by time (d=v⋅td = v \cdot t), so time equals distance divided by speed.
2
Formulate the algebraic equation using the given difference in flight durations.
\frac{600}{v - 50} - \frac{600}{v} = 2
The return flight took 22 hours longer than the outbound flight.
3
Clear denominators and simplify into standard quadratic form.
v^2 - 50v - 15,000 = 0
Multiplying both sides by v(v−50)v(v - 50) yields 600v−600v+30,000=2(v2−50v)600v - 600v + 30,000 = 2(v^2 - 50v), which simplifies to 2v2−100v−30,000=02v^2 - 100v - 30,000 = 0 or v2−50v−15,000=0v^2 - 50v - 15,000 = 0.
4
Solve the quadratic equation for vv.
v = 150
Factoring (v−150)(v+100)=0(v - 150)(v + 100) = 0 gives solutions v=150v = 150 or v=−100v = -100. Physical speed must be positive.

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