In a race of 1000 m, A beats B by 100 m or 10 seconds. If they start a race of 1000 m simultaneously from the same point and if B gets injured after running 50 m less than half the race length and due to which his speed gets halved, then by how much time will A beat B?
65 seconds
Understanding the Race Problem
This problem involves calculating the time difference between two runners, A and B, in a 1000 m race under a specific condition where runner B's speed changes mid-race due to an injury. We are given initial information from a different race that helps us determine the speeds of A and B.
Initial Race Analysis (1000 m)
In the first scenario, A beats B by 100 m or 10 seconds in a 1000 m race. This tells us two important things:
- When A finishes the 1000 m race, B is 100 m behind, meaning B has run 1000 m - 100 m = 900 m.
- The time difference between A finishing and B finishing is 10 seconds. Since A beats B, B takes 10 seconds more than A to complete the 1000 m race.
From the fact that B is 100 m behind when A finishes, and this distance represents a time difference of 10 seconds, we can determine B's speed.
Calculating Speeds
B runs the distance of 100 m (the distance by which A beats B) in 10 seconds. Therefore, B's speed (\(v_B\)) can be calculated as:
\( v_B = \frac{\text{Distance}}{\text{Time}} \)
\( v_B = \frac{100 \text{ m}}{10 \text{ s}} = 10 \text{ m/s} \)
Now we can find the time B takes to complete the 1000 m race at this speed:
\( \text{Time taken by B for 1000 m} = \frac{\text{Distance}}{\text{Speed}} \)
\( \text{Time taken by B for 1000 m} = \frac{1000 \text{ m}}{10 \text{ m/s}} = 100 \text{ seconds} \)
Since A beats B by 10 seconds, A takes 10 seconds less than B to complete the 1000 m race:
\( \text{Time taken by A for 1000 m} = \text{Time taken by B for 1000 m} - 10 \text{ seconds} \)
\( \text{Time taken by A for 1000 m} = 100 \text{ s} - 10 \text{ s} = 90 \text{ seconds} \)
We can also calculate A's speed (\(v_A\)), although it's not strictly necessary for the final answer, it's good to know:
\( v_A = \frac{1000 \text{ m}}{90 \text{ s}} = \frac{100}{9} \text{ m/s} \)
Second Race Scenario: B's Injury
Now consider the second race, also 1000 m, where B gets injured. The injury occurs after B runs a distance that is "50 m less than half the race length".
Half the race length is \( \frac{1000 \text{ m}}{2} = 500 \text{ m} \).
The distance B runs before getting injured is \( 500 \text{ m} - 50 \text{ m} = 450 \text{ m} \).
For the first 450 m, B runs at his original speed of 10 m/s.
After running 450 m, B's speed gets halved. His new speed becomes \( \frac{10 \text{ m/s}}{2} = 5 \text{ m/s} \).
The remaining distance for B to run is \( 1000 \text{ m} - 450 \text{ m} = 550 \text{ m} \).
Calculating B's Total Time in the Second Race
B's total time in the second race is the sum of the time taken for the first 450 m and the time taken for the remaining 550 m.
Time for the first 450 m (at 10 m/s):
\( \text{Time}_1 = \frac{450 \text{ m}}{10 \text{ m/s}} = 45 \text{ seconds} \)
Time for the remaining 550 m (at 5 m/s):
\( \text{Time}_2 = \frac{550 \text{ m}}{5 \text{ m/s}} = 110 \text{ seconds} \)
B's total time for the 1000 m race is:
\( \text{Total Time}_B = \text{Time}_1 + \text{Time}_2 = 45 \text{ s} + 110 \text{ s} = 155 \text{ seconds} \)
Calculating A's Time in the Second Race
A runs the 1000 m race at his constant speed (\(v_A = \frac{100}{9} \text{ m/s}\)). His time for the 1000 m race remains the same as calculated from the first scenario.
\( \text{Total Time}_A = 90 \text{ seconds} \)
Finding the Time Difference
To find by how much time A beats B, we subtract A's total time from B's total time:
\( \text{Time Difference} = \text{Total Time}_B - \text{Total Time}_A \)
\( \text{Time Difference} = 155 \text{ s} - 90 \text{ s} = 65 \text{ seconds} \)
Therefore, A will beat B by 65 seconds in the second race.
| Runner | Speed (m/s) | Distance Run Before Injury (m) | Time Before Injury (s) | Distance Run After Injury (m) | Injured Speed (m/s) | Time After Injury (s) | Total Time (s) |
|---|---|---|---|---|---|---|---|
| A | \( \frac{100}{9} \) | 1000 | 90 | N/A | N/A | N/A | 90 |
| B | 10 | 450 | \( \frac{450}{10} = 45 \) | 550 | 5 | \( \frac{550}{5} = 110 \) | \( 45 + 110 = 155 \) |
Revision Table: Key Calculations
| Concept | Calculation | Value |
|---|---|---|
| B's speed (\(v_B\)) | \( \frac{100 \text{ m}}{10 \text{ s}} \) | 10 m/s |
| Time for B to run 1000m | \( \frac{1000 \text{ m}}{10 \text{ m/s}} \) | 100 seconds |
| Time for A to run 1000m (\(\text{Total Time}_A\)) | 100 s - 10 s | 90 seconds |
| Distance B runs before injury | \( \frac{1000}{2} - 50 \) | 450 m |
| Time for B to run 450m | \( \frac{450 \text{ m}}{10 \text{ m/s}} \) | 45 seconds |
| B's injured speed | \( \frac{10 \text{ m/s}}{2} \) | 5 m/s |
| Distance B runs after injury | 1000 m - 450 m | 550 m |
| Time for B to run 550m (injured) | \( \frac{550 \text{ m}}{5 \text{ m/s}} \) | 110 seconds |
| B's total time (\(\text{Total Time}_B\)) | 45 s + 110 s | 155 seconds |
| Time A beats B by | \( \text{Total Time}_B - \text{Total Time}_A \) | 155 s - 90 s = 65 seconds |
Additional Information on Race Problems
Race problems often test your understanding of speed, time, and distance relationships. Key concepts include:
- Speed, Distance, Time Formula: \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \). This can be rearranged to find distance (\( \text{Distance} = \text{Speed} \times \text{Time} \)) or time (\( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)).
- Head Start: A head start can be given in terms of distance (e.g., A gives B a 10 m head start) or time (e.g., A gives B a 5 second head start). This affects the distance or time each person needs to cover relative to the other.
- Beating by Distance/Time: If A beats B by 'x' meters or 't' seconds in a race of 'D' meters:
- When A covers D meters, B covers (D-x) meters.
- A takes 'T' time to finish, B takes (T+t) time to finish.
- The distance 'x' is covered by B in time 't'. This can be used to find B's speed (\( v_B = x/t \)).
- Relative Speed: Sometimes, thinking about the difference in speeds can simplify problems, especially when dealing with varying distances or times.
Complex problems, like this one, involve combining these concepts and sometimes breaking the race into different segments with different speeds.
Similar Questions
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Important Questions from Relative Speed
In a circular race of 2500 m, a man and a woman start from a point towards opposite directions with speeds of 37 km/h and 35 km/h, respectively. After how much time from the start of the race will they meet for the first time?
- X and Yrun a 3 km race along a circular course of length 300 m. Their speeds are in the ratio 3 : 2. If they start together in the same direction, how many times would the first one pass the other (the start-off is not counted as passing)?
A train of length 600 meters passes another train of length 1000 meters moving in the opposite direction in 96 seconds. If the speed of both the trains is the same, then what is the speed of each train?
The distance between Delhi and Patna is $480$ km. A train starts from Delhi and travels towards Patna at a speed of $60$ kmph. Another train starts from Patna towards Delhi at a speed of $80$ kmph, but starts $1$ hour after the first train. In how much time, after the first train started, will they meet each other?
Ram traveling at the speed of 3 km/h reaches 15 minutes late. Had he walked at 4 km/h, he would have reached 15 minutes earlier. How much distance does Ram have to cover?
