Solving the Racer Speed and Time Difference Problem

This problem involves two racers running the same distance but at different speeds, resulting in different times taken. We are given their speeds and the difference in their finish times. Our goal is to find the total distance of the race.

Understanding the Concepts: Speed, Distance, and Time

The fundamental relationship between speed, distance, and time is:

$$ \text{Speed} = \frac{\text{Distance}}{\text{Time}} $$

From this, we can also express time and distance:

• $$ \text{Time} = \frac{\text{Distance}}{\text{Speed}} $$
• $$ \text{Distance} = \text{Speed} \times \text{Time} $$

In this problem, the distance is constant for both racers, but their speeds and times vary.

Setting up the Problem

Let's define the variables:

• Let the total distance of the race be \(D\) meters.
• Speed of the first racer (\(S_1\)) = 100 m/min.
• Speed of the second racer (\(S_2\)) = 120 m/min.
• Let the time taken by the first racer be \(T_1\) minutes.
• Let the time taken by the second racer be \(T_2\) minutes.

Using the formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \), we can write the times for each racer:

• Time taken by the first racer: $$ T_1 = \frac{D}{S_1} = \frac{D}{100} \text{ minutes} $$
• Time taken by the second racer: $$ T_2 = \frac{D}{S_2} = \frac{D}{120} \text{ minutes} $$

We are told that the second racer takes 10 minutes less than the first racer. This gives us the equation relating their times:

$$ T_1 - T_2 = 10 \text{ minutes} $$

Solving for the Distance

Now, substitute the expressions for \(T_1\) and \(T_2\) into the time difference equation:

$$ \frac{D}{100} - \frac{D}{120} = 10 $$

To solve for \(D\), we need to find a common denominator for the fractions. The least common multiple (LCM) of 100 and 120 is 600.

Multiply the entire equation by 600 to eliminate the denominators:

$$ 600 \times \left( \frac{D}{100} - \frac{D}{120} \right) = 600 \times 10 $$

$$ \left( 600 \times \frac{D}{100} \right) - \left( 600 \times \frac{D}{120} \right) = 6000 $$

$$ 6D - 5D = 6000 $$

Simplify the equation:

$$ D = 6000 \text{ meters} $$

Converting Units

The distance is calculated in meters. The options provided are in kilometers. We need to convert meters to kilometers. Remember that 1 kilometer = 1000 meters.

So, to convert meters to kilometers, we divide by 1000:

$$ D = \frac{6000 \text{ meters}}{1000 \text{ meters/km}} = 6 \text{ kilometers} $$

The length of the race is 6 kilometers.

Verification

Let's check if this distance satisfies the conditions:

Distance = 6000 meters.

• Time for the first racer \( T_1 = \frac{6000 \text{ m}}{100 \text{ m/min}} = 60 \text{ minutes} \).
• Time for the second racer \( T_2 = \frac{6000 \text{ m}}{120 \text{ m/min}} = 50 \text{ minutes} \).

The difference in time is \( T_1 - T_2 = 60 \text{ min} - 50 \text{ min} = 10 \text{ minutes} \). This matches the information given in the problem, confirming our calculation is correct.

Revision Table: Key Concepts for Race Problems

Additional Information: Relative Speed

While not directly used in this specific problem (where racers cover the same distance separately), the concept of relative speed is important for related problems, such as when objects move towards or away from each other, or on a circular track.

• Objects moving in the same direction: Relative speed is the difference between their speeds (\( S_{relative} = |S_1 - S_2| \)). This is useful for calculating the time it takes for one object to overtake another.
• Objects moving in opposite directions: Relative speed is the sum of their speeds (\( S_{relative} = S_1 + S_2 \)). This is useful for calculating the time it takes for them to meet.

Understanding these variations helps tackle a wider range of speed, distance, and time problems in competitive exams.