Question:
A person is running with a uniform velocity towards a flyover. He takes 5 s to reach the flyover from a reference point and takes 50 s to cross the flyover from the same reference point. If the length of the flyover is 1000 m, then his velocity is nearly:
A person is running with a uniform velocity towards a flyover. He takes 5 s to reach the flyover from a reference point and takes 50 s to cross the flyover from the same reference point. If the length of the flyover is 1000 m, then his velocity is nearly:
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In relative motion or multi-segment motion problems, always split the journey into parts and write separate distance-time equations for each segment before solving.
Updated On: Jul 18, 2026
- 83.1 kmph
- 80.0 kmph
- 75.4 kmph
- 85.2 kmph
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the physical situation clearly.
The person is moving with a uniform velocity \(v\) towards a flyover. A reference point is fixed, and two time measurements are given: first, the time taken to reach the flyover (5 s), and second, the time taken to cross the entire flyover (50 s). This indicates a distance-time relation problem involving uniform motion, where speed remains constant throughout the journey.
Step 2: Interpreting the first condition (reaching the flyover).
Let the distance between the reference point and the start of the flyover be \(d\). Since the person takes 5 seconds to reach the flyover, we write: \[ d = v \times 5 \] This expresses the first relation between distance, velocity, and time. So, \[ d = 5v \] This will later be used in forming the total displacement equation.
Step 3: Interpreting the second condition (crossing the flyover).
To completely cross the flyover, the person must cover: - distance \(d\) (from reference point to flyover start), and - length of flyover \(1000 \, m\). So total distance covered in 50 seconds is: \[ d + 1000 = v \times 50 \] Substituting \(d = 5v\), we get: \[ 5v + 1000 = 50v \] This equation now connects all given parameters.
Step 4: Solving for velocity.
Rearranging the equation: \[ 1000 = 50v - 5v \] \[ 1000 = 45v \] \[ v = \frac{1000}{45} = 22.22 \, \text{m/s (approx)} \] This gives the velocity in SI units (m/s), which is not yet in kmph.
Step 5: Converting velocity into kmph.
We use the standard conversion: \[ 1 \, \text{m/s} = 3.6 \, \text{kmph} \] So, \[ v = 22.22 \times 3.6 = 79.99 \, \text{kmph} \] Thus, \[ v \approx 80.0 \, \text{kmph} \] This matches one of the given options directly.
Step 6: Final verification and conclusion.
The computed velocity satisfies both conditions: - correct time to reach flyover (5 s), and - correct total crossing time (50 s). Hence, the nearest and correct value is: \[ \boxed{80.0 \, \text{kmph}} \]
The person is moving with a uniform velocity \(v\) towards a flyover. A reference point is fixed, and two time measurements are given: first, the time taken to reach the flyover (5 s), and second, the time taken to cross the entire flyover (50 s). This indicates a distance-time relation problem involving uniform motion, where speed remains constant throughout the journey.
Step 2: Interpreting the first condition (reaching the flyover).
Let the distance between the reference point and the start of the flyover be \(d\). Since the person takes 5 seconds to reach the flyover, we write: \[ d = v \times 5 \] This expresses the first relation between distance, velocity, and time. So, \[ d = 5v \] This will later be used in forming the total displacement equation.
Step 3: Interpreting the second condition (crossing the flyover).
To completely cross the flyover, the person must cover: - distance \(d\) (from reference point to flyover start), and - length of flyover \(1000 \, m\). So total distance covered in 50 seconds is: \[ d + 1000 = v \times 50 \] Substituting \(d = 5v\), we get: \[ 5v + 1000 = 50v \] This equation now connects all given parameters.
Step 4: Solving for velocity.
Rearranging the equation: \[ 1000 = 50v - 5v \] \[ 1000 = 45v \] \[ v = \frac{1000}{45} = 22.22 \, \text{m/s (approx)} \] This gives the velocity in SI units (m/s), which is not yet in kmph.
Step 5: Converting velocity into kmph.
We use the standard conversion: \[ 1 \, \text{m/s} = 3.6 \, \text{kmph} \] So, \[ v = 22.22 \times 3.6 = 79.99 \, \text{kmph} \] Thus, \[ v \approx 80.0 \, \text{kmph} \] This matches one of the given options directly.
Step 6: Final verification and conclusion.
The computed velocity satisfies both conditions: - correct time to reach flyover (5 s), and - correct total crossing time (50 s). Hence, the nearest and correct value is: \[ \boxed{80.0 \, \text{kmph}} \]
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