Question

In a flight of $600\mathrm{km}$, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by $200\mathrm{km}/\mathrm{h}$ and the time of flight was increased by $30\min$. Find the duration of flight.

Answer 1

Step 1:

Form a quadratic equation:

Let the original speed of the aircraft be $x\mathrm{km}/\mathrm{h}$.

Given that,

The speed was reduced by $200\mathrm{km}/\mathrm{h}$.

The new speed will be $\left(x-200\right)\mathrm{km}/\mathrm{h}$.

Use the following formula:

$\mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{speed}}$

The duration of flight at original speed will be $\frac{600}{x}\mathrm{h}$.

The duration of flight at a reduced speed will be $\frac{600}{x-200}\mathrm{h}$.

Convert the increased time of flight into hours:

$\begin{array}{rcl}\mathrm{increased}\mathrm{time}\mathrm{of}\mathrm{flight}& =& \frac{30}{60}\mathrm{h}\\ & =& \frac{1}{2}\mathrm{h}\end{array}$

By the given condition,

$\begin{array}{rcl}\frac{600}{x-200}-\frac{600}{x}& =& \frac{1}{2}\\ & \Rightarrow & 600\left(\frac{1}{x-200}-\frac{1}{x}\right)=\frac{1}{2}\\ & \Rightarrow & 600\left(\frac{x-\left(x-200\right)}{x\left(x-200\right)}\right)=\frac{1}{2}\\ & \Rightarrow & 600\left(\frac{x-x+200}{x^2-200x}\right)=\frac{1}{2}\\ & \Rightarrow & \frac{600\times 200}{x^2-200x}=\frac{1}{2}\\ & \Rightarrow & 2\times 600\times 200=x^2-200x\left(\mathrm{On}\mathrm{cross}\mathrm{multiplication}\right)\\ & \Rightarrow & 240000=x^2-200x\\ & \Rightarrow & x^2-200x-240000=0\end{array}$

Step 2:

Solve the quadratic equation:

Using the factorization method:

$\begin{array}{rcl}& \Rightarrow & x^2-600x+400x-240000=0\\ & \Rightarrow & x\left(x-600\right)+400\left(x-600\right)=0\\ & \Rightarrow & \left(x+400\right)\left(x-600\right)=0\\ & \Rightarrow & x+400=0orx-600=0\\ & \Rightarrow & x=-400orx=600\end{array}$

But time cannot be negative, rejecting $x=-400$

So, the original speed of the aircraft was $600\mathrm{km}/\mathrm{h}$.

Therefore,

$\begin{array}{rcl}\mathrm{duration}\mathrm{of}\mathrm{flight}& =& \frac{600}{x}\mathrm{h}\\ & =& \frac{600}{600}\mathrm{h}\\ & =& 1\mathrm{hr}\end{array}$

Final answer:

Hence, the duration of the flight is $1\mathrm{hr}$.