Solving the Train Speed Problem with a Delay

This problem involves a train covering a certain distance, experiencing a technical fault that reduces its speed for a portion of the journey, and consequently getting delayed. We need to find the train's original or normal speed.

Understanding the Train Speed Scenario

Here's a breakdown of the information given:

• Total distance to cover: 370 km
• Initial part of the journey: 100 km covered at the normal speed.
• Remaining part of the journey: 370 km - 100 km = 270 km.
• Speed during the remaining part: Normal speed minus 5 km/h.
• Total delay in reaching the destination: 36 minutes.

The key to solving this problem is to relate the times taken under the normal conditions and the fault conditions to the given delay.

Setting up the Equations for Time and Speed

Let the normal speed of the train be \( v \) km/h.

The standard formula relating distance, speed, and time is: \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \).

Case 1: Normal Conditions (No Fault)

If the train had run at its normal speed \( v \) for the entire 370 km, the normal time taken (\( T_{normal} \)) would be:

\( T_{normal} = \frac{370}{v} \) hours.

Case 2: With Technical Fault

The journey is split into two parts:

1. First 100 km: Speed = \( v \) km/h. Time taken (\( t_1 \)) = \( \frac{100}{v} \) hours.
2. Remaining 270 km: Speed = \( v - 5 \) km/h. Time taken (\( t_2 \)) = \( \frac{270}{v-5} \) hours.

The actual total time taken (\( T_{actual} \)) is the sum of the times for these two parts:

\( T_{actual} = t_1 + t_2 = \frac{100}{v} + \frac{270}{v-5} \) hours.

Using the Delay Information

The train was delayed by 36 minutes. A delay means the actual time taken was more than the normal time. So, the difference between the actual time and the normal time is the delay.

\( T_{actual} - T_{normal} = \text{Delay} \)

First, convert the delay from minutes to hours:

\( \text{Delay} = 36 \text{ minutes} = \frac{36}{60} \text{ hours} = \frac{3}{5} \text{ hours} \).

Now, substitute the expressions for \( T_{actual} \) and \( T_{normal} \) into the delay equation:

\( \left( \frac{100}{v} + \frac{270}{v-5} \right) - \frac{370}{v} = \frac{3}{5} \)

Solving for the Normal Speed \( v \)

Simplify the equation:

\( \frac{100}{v} - \frac{370}{v} + \frac{270}{v-5} = \frac{3}{5} \)

\( \frac{100 - 370}{v} + \frac{270}{v-5} = \frac{3}{5} \)

\( \frac{-270}{v} + \frac{270}{v-5} = \frac{3}{5} \)

Rearrange the terms:

\( \frac{270}{v-5} - \frac{270}{v} = \frac{3}{5} \)

Find a common denominator on the left side, which is \( v(v-5) \):

\( \frac{270v - 270(v-5)}{v(v-5)} = \frac{3}{5} \)

\( \frac{270v - 270v + 1350}{v(v-5)} = \frac{3}{5} \)

\( \frac{1350}{v(v-5)} = \frac{3}{5} \)

Cross-multiply:

\( 5 \times 1350 = 3 \times v(v-5) \)

\( 6750 = 3v(v-5) \)

\( 6750 = 3v^2 - 15v \)

Divide the entire equation by 3:

\( \frac{6750}{3} = \frac{3v^2}{3} - \frac{15v}{3} \)

\( 2250 = v^2 - 5v \)

Rearrange into a standard quadratic equation form \( av^2 + bv + c = 0 \):

\( v^2 - 5v - 2250 = 0 \)

Solving the Quadratic Equation

We need to find the values of \( v \) that satisfy this equation. We can factor the quadratic equation. We look for two numbers that multiply to -2250 and add up to -5. These numbers are 45 and -50.

So, the equation can be factored as:

\( (v + 45)(v - 50) = 0 \)

This gives two possible solutions for \( v \):

• \( v + 45 = 0 \implies v = -45 \)
• \( v - 50 = 0 \implies v = 50 \)

Since speed cannot be negative, the normal speed of the train must be \( v = 50 \) km/h.

Verification of the Normal Speed

Let's check if a normal speed of 50 km/h results in a 36-minute delay.

• Normal speed: 50 km/h
• Speed after fault: \( 50 - 5 = 45 \) km/h

Normal time for 370 km at 50 km/h: \( \frac{370}{50} = 7.4 \) hours.

Time taken for the first 100 km at 50 km/h: \( \frac{100}{50} = 2 \) hours.

Time taken for the remaining 270 km at 45 km/h: \( \frac{270}{45} = 6 \) hours.

Actual total time: \( 2 \text{ hours} + 6 \text{ hours} = 8 \) hours.

Delay = Actual time - Normal time = 8 hours - 7.4 hours = 0.6 hours.

Converting the delay back to minutes: \( 0.6 \times 60 = 36 \) minutes.

This matches the delay given in the problem statement. Therefore, the normal speed of the train is 50 km/h.

The relationship \( \left( \frac{100}{v} + \frac{270}{v-5} \right) - \frac{370}{v} = \frac{3}{5} \) is the core equation derived from the delay.

Revision Table: Key Concepts

Additional Information: Solving Speed, Distance, and Time Problems

Speed, distance, and time problems often involve setting up equations based on the relationships between these quantities. When conditions change (like a change in speed or route), it affects the time taken. Delays or early arrivals indicate a difference between the actual time and the planned time.

Steps typically involve:

• Identifying the known and unknown variables (usually speed, distance, or time).
• Setting up equations based on the information given for different parts of the journey or different scenarios (e.g., normal vs. changed speed).
• Using any given relationship between times (like delay or difference) to form a final equation.
• Solving the equation(s) to find the unknown variable.

These problems can sometimes lead to linear or quadratic equations, as seen in this train speed example.