Two runners in a 100-m race start from the same place. Runner A starts as soon as the starting gun is fired and runs at a constant speed of 8.00 m/s Runner B starts 2.00 s later and runs at a constant speed of 10 m/s :-

Options

A. If F=50N ,`theta =45^(@)`
B. If F=50N ,`theta=tan^(-)(1//2)`
C. If F=0,`theta=0^(@)`
D. If F = 100N, `theta = 45^(@)`

Answer

To solve the problem of two runners in a 100-m race, we need to determine when Runner A and Runner B meet, given their speeds and the delay in Runner B's start.

### Step-by-Step Solution:

1. **Identify the speeds and delay:**
- Runner A's speed = 8.00 m/s
- Runner B's speed = 10.00 m/s
- Delay for Runner B = 2.00 s

2. **Calculate the distance covered by Runner A before Runner B starts:**
- In 2 seconds, Runner A will cover:
\[ \text{Distance} = \text{Speed} \times \text{Time} = 8.00 \, \text{m/s} \times 2.00 \, \text{s} = 16.00 \, \text{m} \]

3. **Determine the distance remaining for Runner A when Runner B starts:**
- Total distance of the race = 100 m
- Distance remaining for Runner A when Runner B starts:
\[ 100 \, \text{m} - 16.00 \, \text{m} = 84.00 \, \text{m} \]

4. **Set up the equation for the time taken for both runners to finish the race:**
- Let \( t \) be the time in seconds that Runner B runs after starting.
- The distance Runner B covers in time \( t \) is:
\[ \text{Distance} = \text{Speed} \times \text{Time} = 10.00 \, \text{m/s} \times t \]
- The distance Runner A covers in the same time \( t \) (after Runner B starts) is:
\[ \text{Distance} = 8.00 \, \text{m/s} \times (t + 2) \]

5. **Set the distances equal to each other when they meet:**
- Since both runners will have covered the same distance when they meet:
\[ 10t = 16 + 8(t + 2) \]

6. **Simplify the equation:**
\[ 10t = 16 + 8t + 16 \]
\[ 10t = 8t + 32 \]
\[ 10t - 8t = 32 \]
\[ 2t = 32 \]
\[ t = 16 \, \text{s} \]

7. **Calculate the total time for Runner A:**
- Total time for Runner A from the start until they meet:
\[ \text{Total time} = t + 2 = 16 + 2 = 18 \, \text{s} \]

8. **Verify the distances:**
- Distance covered by Runner A in 18 seconds:
\[ \text{Distance} = 8.00 \, \text{m/s} \times 18 \, \text{s} = 144 \, \text{m} \]
- Distance covered by Runner B in 16 seconds:
\[ \text{Distance} = 10.00 \, \text{m/s} \times 16 \, \text{s} = 160 \, \text{m} \]
- Since Runner A runs 84 m after Runner B starts, we can confirm they meet at the same point.

### Conclusion:
Both runners meet after 18 seconds from the start of Runner A.