Question

In a flight of 1600 km an aircraft was slowed down due to bad weather. Its speed was reduced by 400 kilometre per hour and hence the time of flight increased by 40 minutes. Find the duration of the aircraft​

Answer 1

Answer:

ANSWER

In a flight of 6000 km, an aircraft was slowed down due to bad weather.

Its average speed for the trip was reduced by 400km/h and the time increased by 30 min.

Find the original duration of the flight

:

The weather has to be really bad to slow the plane down 400 km/hr, anyway:

:

Let s = original effective speed

then

(s-400) = speed in bad weather

:

Write a time equation; time = dist/speed

Bad weather time - normal time = 1/2 hr

6000%2F%28%28s-400%29%29 - 6000%2Fs = .5

multiply by s(s-400) to clear the denominators; results:

6000s - 6000(s-400) = .5s(s-400)

6000s - 6000s + 2400000 = .5s^2 - 200s

A quadratic equation

.5s^2 - 200s - 2400000

solve this using the quadratic formula

You should get a positive solution of:

s = 2400 km/hr is the normal speed

:

Find the duration of the flight

6000/2400 = 2.5 hrs

Check this by finding the duration when speed reduced by 400 km/hr

6000/2000 = 3 hrs, a half hour longer

Answer 2

Answer:

2.5 hours

Step-by-step explanation:

Let the original speed of the aircraft be 'x' km/hr Decreased speed of the aircraft = (x - 400) km/hr Time = Distance/Speed Original duration of the flight = (6000 / x) hrs Duration of the flight with decreased speed = (6000 / x-400) hrs Difference of time = 30 min = 1/2 hrs⇒ (6000 / x-400) - (6000 / x) = 1/2 ⇒ (6000x - 6000x + 2400000) / x(x - 400) = 1/2 ⇒ x^2 - 400x = 4800000 ⇒ x^2 - 400x - 4800000 = 0 ⇒ (x - 2400)(x + 2000) = 0 ⇒ x = 2400 or -2000Positive value of x is 2400. Therefore original speed of the aircraft is 2400 km/hr. Original duration of the aircraft = 6000/2400 = 2.5 hrs.

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