Understanding the Race Problem with Speed and Distance

This problem involves two runners, Aravind and Bhanu, with different speeds. Aravind runs faster than Bhanu. Bhanu is given a head start, and we need to find the distance from the starting point where Aravind catches up to Bhanu.

Analyzing the Given Information

• Aravind's speed is \( \frac{5}{4} \) times Bhanu's speed. This means the ratio of their speeds is \( S_{Aravind} : S_{Bhanu} = \frac{5}{4} : 1 = 5 : 4 \).
• Aravind gives Bhanu a lead of 60 m. This means Bhanu starts 60 meters ahead of Aravind.

Relating Speed, Distance, and Time

When Aravind catches up to Bhanu, both runners will have been running for the same amount of time. The fundamental relationship between speed, distance, and time is:

\( \text{Distance} = \text{Speed} \times \text{Time} \)

Since the time is the same for both runners until they meet, the ratio of the distances they cover will be equal to the ratio of their speeds.

\( \frac{\text{Distance covered by Aravind}}{\text{Distance covered by Bhanu}} = \frac{\text{Speed of Aravind}}{\text{Speed of Bhanu}} = \frac{5}{4} \)

Setting up the Distances

Let \( D \) be the distance from the starting point where Aravind catches up to Bhanu. This is the total distance covered by Aravind.

• Distance covered by Aravind = \( D \) meters.
• Since Bhanu started 60 m ahead, the distance covered by Bhanu when they meet at point \( D \) from the start is \( D - 60 \) meters.

Using the Ratio of Distances

We know the ratio of distances covered is 5:4. So, we can write the equation:

\( \frac{D}{D - 60} = \frac{5}{4} \)

Solving for the Distance (D)

Now, we solve this equation for \( D \):

1. Cross-multiply: \( 4 \times D = 5 \times (D - 60) \)
2. Distribute the 5: \( 4D = 5D - 5 \times 60 \)
3. Simplify: \( 4D = 5D - 300 \)
4. Subtract \( 4D \) from both sides: \( 0 = 5D - 4D - 300 \)
5. Simplify: \( 0 = D - 300 \)
6. Add 300 to both sides: \( D = 300 \)

So, Aravind catches up to Bhanu at a distance of 300 meters from the starting point.

Verification

If they meet at 300 m from the start:

• Aravind covers 300 m.
• Bhanu covers \( 300 - 60 = 240 \) m.
• The ratio of distances covered is \( 300 : 240 \).
• Dividing both by 60, we get \( \frac{300}{60} : \frac{240}{60} = 5 : 4 \).

This ratio matches the given speed ratio of 5:4, confirming our answer.

Revision Table: Key Concepts

Additional Information: Relative Speed in Races

In problems like this, the concept of relative speed is useful, although the ratio method works perfectly here. Relative speed is the speed of one object with respect to another.

• When two objects move in the same direction, their relative speed is the difference between their speeds. In this case, Aravind is catching up to Bhanu, so the relative speed at which Aravind gains on Bhanu is \( S_{Aravind} - S_{Bhanu} \).
• Let's say Bhanu's speed is \( S \). Then Aravind's speed is \( \frac{5}{4}S \). The relative speed is \( \frac{5}{4}S - S = (\frac{5}{4} - 1)S = \frac{1}{4}S \).
• Aravind needs to cover the initial gap of 60 m at this relative speed.
• Time taken to cover the gap = \( \frac{\text{Initial Gap}}{\text{Relative Speed}} = \frac{60}{\frac{1}{4}S} = \frac{240}{S} \) seconds (or units of time).
• In this time, Aravind covers a distance \( D \). \( D = S_{Aravind} \times \text{Time} = \frac{5}{4}S \times \frac{240}{S} \).
• The \( S \) terms cancel out: \( D = \frac{5}{4} \times 240 \).
• \( D = 5 \times \frac{240}{4} = 5 \times 60 = 300 \) m.

Both the ratio method and the relative speed method yield the same result, demonstrating different ways to approach time and distance problems involving head starts.