Question

In a flight of 6000km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 400 km/hr and time increased by 30 minutes. Find the original duration of the flight.

Answer 1

Hint: In this question let the original average speed of the aircraft be V km/hr and time taken be t hours. Use the constraints of the questions to formulate equations by using the relationship between distance, speed and time.

Complete step-by-step answer:
Let the average speed of the aircraft be V km/hr.
And the time it takes to cover a distance be (t) hr.
Now it is given that the flight covers a distance (d) = 6000 km with some speed.

Now it is given that its average speed is decreased by 400 km/hr and it will cover the same distance in 30 minutes more.
Now as we know 60 min = 1 hour.
Therefore 30 min = (30/60) = (1/2) hour
Therefore new speed = (V – 400) km/hr.
And the new time = (t + (1/2)) hr.
Now the distance is the same in both the cases.
And we know that the distance, speed and time is related as
Distance = speed $ \times $ time.
Therefore
$ \Rightarrow 6000 = V.t$ Km............................. (1)
And
\[ \Rightarrow 6000 = \left( {V - 400} \right)\left( {t + \dfrac{1}{2}} \right)\] Km.............................. (2)
So equate these two equation we have,
\[ \Rightarrow \left( {V - 400} \right)\left( {t + \dfrac{1}{2}} \right) = V.t\]
Now simplify the above equation we have,
\[ \Rightarrow V.t + \dfrac{1}{2}V - 400t - 200 = V.t\]
\[ \Rightarrow \dfrac{1}{2}V - 400t - 200 = 0\]

Now from equation (1) \[V = \dfrac{{6000}}{t}\] so substitute this value in above equation we have,
\[ \Rightarrow \dfrac{1}{2}.\dfrac{{6000}}{t} - 400t - 200 = 0\]
Now simplify the above equation we have,
\[ \Rightarrow 3000 - 400{t^2} - 200t = 0\]
Divide by -200 throughout we have,
\[ \Rightarrow 2{t^2} + t - 15 = 0\]
Now factorize this equation we have,
\[ \Rightarrow 2{t^2} + 6t - 5t - 15 = 0\]
$ \Rightarrow 2t\left( {t + 3} \right) - 5\left( {t + 3} \right) = 0$
$ \Rightarrow \left( {t + 3} \right)\left( {2t - 5} \right) = 0$
$ \Rightarrow t = \dfrac{5}{2}, - 3$
Negative time is not possible.
So the original duration of the flight is (5/2) = 2.5 hr.
So this is the required answer.

Note: In this firstly we have calculated the new decreased speed as (v-400) km/hr and the new time as $\left( {t + \dfrac{1}{2}} \right)$hrs as the plane got delayed by 30 mins. Now originally if the plane would have been travelling at v km/hr taking t hours’ time then it will be traveling 6000km only, and now with the new speed and the new timings the distance travelled has to be the same. This concept helps solve it.