Question

Two planes left simultaneously from the same airport and headed in the same direction towards another airport 3600 km away. The speed of one of the planes was 200 km/hour slower than the speed of the other plane, and so it arrived at its destination 1.5 hours after the faster plane. Find the speeds of both planes.

Answer 1

Answer:

• 800 kph
• 600 kph

Step-by-step explanation:

Equation

Let s represent the speed of the faster plane. Then its travel time is ...

  time = distance/speed = 3600/s

travel time for the slower plane is ...

  time = distance/speed = 3600/(s -200)

The difference in these times is 1.5 hours, so we have ...

  3600/(s -200) -3600/s = 1.5

Solution

Multiplying by (2/3)s(s -200), we get ...

  2400(s) -2400(s -200) = s(s -200)

  s^2 -200s -480000 = 0 . . . . . . . . . . rewrite in standard form

  (s -800)(s +600) = 0 . . . . . . . . . . . . . factor

The positive value of s that makes a factor zero is s = 800.

Conclusion

The speed of the faster plane was 800 km/h; of the slower plane, 600 km/h.

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Check

The travel time for the faster plane was 3600/800 = 4.5 hours. For the slower plane, 3600/600 = 6 hours, a difference of 1.5 hours.