In a 700 m race, Amit reaches the finish point in 20 seconds and Rahul reaches in 25 seconds. Amit beats Rahul by a distance of :
140 m
Understanding Race Dynamics: Calculating Distance Difference
This question asks us to determine the distance by which Amit beats Rahul in a 700 m race, given their finishing times.
In a race, when one person 'beats' another by a certain distance, it means that when the first person (the winner) finishes the race, the second person is still that calculated distance away from the finish line.
Step-by-Step Solution to the Race Problem
First, let's find the speed of each runner. Speed is calculated as Distance divided by Time.
1. Calculate Amit's Speed
Amit completes the 700 m race in 20 seconds.
Using the formula: $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$
Amit's speed $= \frac{700 \text{ m}}{20 \text{ s}} = 35 \text{ m/s}$
2. Calculate Rahul's Speed
Rahul completes the 700 m race in 25 seconds.
Using the formula: $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$
Rahul's speed $= \frac{700 \text{ m}}{25 \text{ s}} = 28 \text{ m/s}$
3. Determine Rahul's Position When Amit Finishes
Amit finishes the race in 20 seconds. In these 20 seconds, Rahul, running at his speed, covers a certain distance.
Using the formula: $\text{Distance} = \text{Speed} \times \text{Time}$
Distance covered by Rahul in 20 seconds $= \text{Rahul's speed} \times 20 \text{ s}$
Distance covered by Rahul in 20 seconds $= 28 \text{ m/s} \times 20 \text{ s} = 560 \text{ m}$
4. Calculate the Distance Amit Beats Rahul By
When Amit reaches the 700 m finish line (at 20 seconds), Rahul has only covered 560 m. The distance by which Amit beats Rahul is the remaining distance Rahul needs to cover to reach the finish line.
Distance Amit beats Rahul by $= \text{Total race distance} - \text{Distance Rahul covered in 20 seconds}$
Distance Amit beats Rahul by $= 700 \text{ m} - 560 \text{ m} = 140 \text{ m}$
Alternatively, we can consider the time difference.
Amit finishes 25 s - 20 s = 5 seconds before Rahul.
In these 5 seconds, Rahul would still be running. The distance Rahul covers in these extra 5 seconds (or the distance he still needs to cover when Amit finishes) is the distance by which Amit beats him.
Distance Rahul covers in 5 seconds $= \text{Rahul's speed} \times \text{Time difference}$
Distance Rahul covers in 5 seconds $= 28 \text{ m/s} \times 5 \text{ s} = 140 \text{ m}$
Both methods give the same result: Amit beats Rahul by a distance of 140 m.
Summary of Calculations
- Amit's Speed: $\frac{700}{20} = 35$ m/s
- Rahul's Speed: $\frac{700}{25} = 28$ m/s
- Distance covered by Rahul in 20 seconds: $28 \times 20 = 560$ m
- Distance difference at 20 seconds: $700 - 560 = 140$ m
- Time difference: $25 - 20 = 5$ seconds
- Distance Rahul covers in 5 seconds: $28 \times 5 = 140$ m
Thus, Amit beats Rahul by 140 m in the 700 m race.
| Metric | Amit | Rahul |
|---|---|---|
| Race Distance | 700 m | 700 m |
| Time Taken | 20 s | 25 s |
| Speed | 35 m/s | 28 m/s |
| Distance at 20s | 700 m (Finished) | 560 m |
| Time Difference | - | 5 s |
Revision Table: Key Concepts in Race Problems
| Concept | Explanation | Formula |
|---|---|---|
| Speed | How fast an object is moving. | $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$ |
| Distance | The total length covered by an object. | $\text{Distance} = \text{Speed} \times \text{Time}$ |
| Time | The duration of the movement. | $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$ |
| Beating by Distance | The distance the slower competitor is from the finish line when the faster competitor finishes. Calculated as Total Distance - Distance covered by Slower runner in the Faster runner's time. Or, Speed of Slower runner × Time difference. | $D_{beat} = D_{total} - (S_{slower} \times T_{faster})$ or $D_{beat} = S_{slower} \times (T_{slower} - T_{faster})$ |
Additional Information on Race Calculations
Race problems often involve calculating speeds, times, or distances based on the principle of uniform motion. It's important to be consistent with units (meters for distance, seconds for time, meters per second for speed).
- When comparing runners, you can either calculate the distance covered by the slower runner in the time the faster runner takes, or calculate the distance the slower runner covers in the time difference between them. Both methods should yield the same result.
- Understanding the relationship between speed, distance, and time is fundamental to solving these problems. The core formula $\text{Distance} = \text{Speed} \times \text{Time}$ can be rearranged to find any of the three variables if the other two are known.
- These concepts are applicable in various real-world scenarios, not just sports races, but also in calculating travel times, distances, and speeds for vehicles, etc.
Important Questions from Linear Programming
The corner points of the feasible region determined by:
x + y ≤ 8, 2x + y ≥ 8, x ≥ 0, y ≥ 0
are A(0, 8), B(4, 0), and C(8, 0). If the objective function Z = ax + by has its maximum value on the line segment AB, then the relation between a and b is:
An objective function Z = ax + by is maximum at points (8, 2) and (4, 6). If a ≥ 0, b ≥ 0, and ab = 25, then the maximum value of the function is:
Which of the following cannot be the direction ratios of the straight line:
\( \frac{x - 3}{2} = \frac{2 - y}{3} = \frac{z + 4}{-1} \)
Corner points of a feasible bounded region are (0,10), (4,2), (3,7) and (10,6). Maximum value 50 of objective function \( z = ax + by \) occurs at two points (0,10) and (10,6). The value of \( a \) and \( b \) are:
The corner points of the feasible region determined by system of linear constraints are (60,0), (120,0), (40,20), and (60,30). Let \( z = ax + by \), where \( a, b > 0 \) be the objective function. Find condition on \( a \) and \( b \) so that the maximum of \( z \) occurs at (120,0) and (60,30).
