Q.10 A plane left 40 minutes late due to bad weather and in order to reach the destination, 1600 km away in time, it had to increase its speed by 400 km/hr from its usual speed. Find the usual speed of the plane.


Let's denote the usual speed of the plane as \( v \) km/hr.

### Step 1: Calculate the time it would take to travel 1600 km at the usual speed
The time taken to travel 1600 km at the usual speed \( v \) is given by:
\[
\text{Time at usual speed} = \frac{1600}{v} \text{ hours}
\]

### Step 2: Determine the time the plane actually had due to the delay
Since the plane left 40 minutes late, it had 40 minutes less to reach the destination on time. 40 minutes is equivalent to \( \frac{40}{60} = \frac{2}{3} \) hours.

So, the time the plane actually had is:
\[
\text{Available time} = \frac{1600}{v} - \frac{2}{3} \text{ hours}
\]

### Step 3: Set up the equation with the increased speed
When the plane increases its speed by 400 km/hr, its new speed becomes \( v + 400 \) km/hr. The time taken to travel 1600 km at this increased speed is:
\[
\text{Time at increased speed} = \frac{1600}{v + 400} \text{ hours}
\]

Since the plane must cover the distance in the reduced time:
\[
\frac{1600}{v + 400} = \frac{1600}{v} - \frac{2}{3}
\]

### Step 4: Solve the equation
Multiply both sides of the equation by \( (v + 400) \times v \) to clear the denominators:
\[
1600v = 1600(v + 400) - \frac{2}{3} v(v + 400)
\]

Expanding and simplifying:
\[
1600v = 1600v + 640000 - \frac{2}{3}v^2 - \frac{800}{3}v
\]
\[
0 = 640000 - \frac{2}{3}v^2 - \frac{800}{3}v
\]

Multiply through by 3 to eliminate the fractions:
\[
0 = 1920000 - 2v^2 - 800v
\]
\[
2v^2 + 800v - 1920000 = 0
\]

### Step 5: Solve the quadratic equation
Divide by 2:
\[
v^2 + 400v - 960000 = 0
\]

Use the quadratic formula \( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = 400 \), and \( c = -960000 \):
\[
v = \frac{-400 \pm \sqrt{400^2 + 4 \times 960000}}{2}
\]
\[
v = \frac{-400 \pm \sqrt{160000 + 3840000}}{2}
\]
\[
v = \frac{-400 \pm \sqrt{4000000}}{2}
\]
\[
v = \frac{-400 \pm 2000}{2}
\]

This gives us two possible solutions:
\[
v = \frac{1600}{2} = 800 \text{ km/hr} \quad \text{or} \quad v = \frac{-2400}{2} = -1200 \text{ km/hr}
\]

Since speed cannot be negative, the usual speed \( v \) is:
\[
\boxed{800 \text{ km/hr}}
\]

### Summary
The usual speed of the plane is 800 km/hr.

Would you like more details or have any questions? Here are some related questions:

1. How do you apply the quadratic formula to solve equations?
2. What are the properties of quadratic equations?
3. How does increasing speed affect travel time?
4. What are the practical applications of such problems in aviation?
5. How can one convert minutes to hours in time calculations?

**Tip:** Always check the units and convert them to match for consistency when solving problems involving time, speed, and distance.

Learn how to solve a speed, distance, time problem involving a plane delayed due to bad weather and the need to increase speed to reach the destination on time. Detailed step-by-step solution provided, including formulas for time, speed, distance, and quadratic equations. Suitable for students in Grades 9-12.

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