a pilot files a distance of 800 km. He saves 40 minutes by increasing the speed by 40km/hr.Find the actual speed of the aircraft using quadratic equation?
Answers
Solution :-
Let us Assume That, actual speed of the aircraft is x km/h.
So,
→ Time in covering a distance of 800km = (Distance/speed) = (800/x) hours.
Now, His Speed is Increased by 40km/h.
So,
→ New Speed of aircraft = (x + 40)km/h.
→ New Time to cover 800km = 800/(x + 40) Hours.
Now we have given That, He saves 40 Minutes.
Therefore,
→ (800/x) - 800/(x + 40) = 40 min = (40/60) = (2/3) Hours
→ 800 [(x+40 -x) /x(x + 40)] = (2/3)
→ 800*40*3 = 2x(x + 40)
→ 48000 = x² + 40x
→ x² + 40x - 48000 = 0
→ x² + 240x - 200x - 48000 = 0
→ x(x + 240) - 200(x + 240) = 0
→ (x + 240) (x - 200) = 0
→ x = (-240) km/h & 200km/h.
Since, speed can't be Negative.
Hence, The Actual Speed of the Aircraft is 200km/h.
We have been given that plot files distance of 800 km. He saves 40 minutes by increasing speed by 40 km/h.
We have to find actual speed of aircraft.
Solution
Let the actual speed of aircraft be y km/h.
We know , distance = Speed × Time
⇒ Time = Distance/Speed
At first, time taken by aircraft = 800/y hours.
Secondly, he saves time = 40 minutes.
⇒ He saves = (40/60) hours.
⇒ He saves = 2/3 hours.
Now, time taken by aircraft in 2nd case = [800/(y + 40)] hours.
According to Question :
⇒ [800/(y + 40)] - [800/y] = 2/3
⇒ [800y - 800(y + 40)]/y(y + 40) = 2/3
⇒ [800y - 800y - 32000]/(y² + 40y) = 2/3
⇒ -32000/(y² + 40y) = 2/3
⇒ 3(-32000) = 2(y² + 40y)
⇒ -96000 = 2y² + 80y
⇒ 2(y² + 40y + 48000 = 0
⇒ y² + 240y - 200y + 48000 = 0
⇒ y(y + 240) - 200(y + 240) = 0
⇒ (y - 200)(y + 240) = 0
⇒ y - 200 = 0 or, y + 240 = 0
⇒ y = 200 km/h or, y = -240 km/h
As, speed of aircraft can't be negative.
Therefore,