Quadratic Equation Word Problem: Speed of a Flight

This lesson works through a quadratic-equation word problem about a flight. A 600 km journey is slowed by bad weather, the average speed drops by 200 km/h, and the trip takes 30 minutes longer. We find the original duration of the flight.

Setting up the problem

Let the original average speed be $x$ kilometres per hour. The distance is fixed:

$\text{distance} = 600 \text{ km}.$

Using $\text{time} = \dfrac{\text{distance}}{\text{speed}}$, the original time is

$t = \frac{600}{x} \text{ hours}.$

Due to bad weather the speed falls by $200$ km/h, so the new speed is $x - 200$ and the new time is

$\frac{600}{x - 200}.$

Translating the condition

The slower flight takes $30$ minutes more than the original. Converting $30$ minutes to hours gives $\tfrac{30}{60} = \tfrac{1}{2}$, so

$\frac{600}{x - 200} = \frac{600}{x} + \frac{1}{2}.$

Combining the fractions

Move $\dfrac{600}{x}$ to the left and factor out $600$:

$600\left(\frac{1}{x - 200} - \frac{1}{x}\right) = \frac{1}{2}.$

Taking the common denominator $x(x - 200)$:

$600 \cdot \frac{x - (x - 200)}{x(x - 200)} = \frac{1}{2}.$

The numerator simplifies, since $x - (x - 200) = 200$:

$600 \cdot \frac{200}{x(x - 200)} = \frac{1}{2}.$

Forming the quadratic

Cross multiplying gives

$600 \cdot 200 \cdot 2 = x(x - 200).$

The left side is $240000$, so

$240000 = x^2 - 200x,$

which rearranges to the standard form

$x^2 - 200x - 240000 = 0.$

Solving by factorisation

We need two numbers whose product is $-240000$ and whose sum is $-200$. The pair $-600$ and $400$ works, since $(-600) \times 400 = -240000$ and $(-600) + 400 = -200$. Choosing signs to match:

$(x - 600)(x + 400) = 0.$

So $x = 600$ or $x = -400$. A speed cannot be negative, so we reject $x = -400$ and keep

$x = 600 \text{ km/h}.$

Finding the original duration

The question asks for the original time, not the speed. Substituting back:

$t = \frac{600}{x} = \frac{600}{600} = 1 \text{ hour}.$

Key takeaways

• Use $\text{distance} = \text{speed} \times \text{time}$, so $\text{time} = \dfrac{\text{distance}}{\text{speed}}$ and $\text{speed} = \dfrac{\text{distance}}{\text{time}}$.
• To convert minutes to hours divide by $60$; to convert hours to minutes multiply by $60$.
• Clearing fractions in a time equation turns the word problem into a quadratic you can solve by factorisation, then reject any root that is physically impossible.