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A train is running at a constant speed o...

A train is running at a constant speed of 90 km/h on a straight track. A person standing at the top of a boggey moves in the direction of motion of the train such that he covers 1 meters on the train each second. The speed of the person with respect to ground to -

A

25 m/s

B

91 km/h

C

26 km/h

D

26 m/s

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the speed of the person with respect to the ground. Here’s a step-by-step breakdown of the solution: ### Step 1: Understand the Given Information - The speed of the train (V_train) is 90 km/h. - The speed of the person relative to the train (V_person/train) is 1 m/s. ### Step 2: Convert the Speed of the Train to m/s Since the speed of the train is given in kilometers per hour, we need to convert it to meters per second for consistency with the speed of the person. - Conversion factor: 1 km/h = (1000 m) / (3600 s) = 5/18 m/s. - Therefore, V_train in m/s = 90 km/h × (5/18) = 25 m/s. ### Step 3: Use the Relative Velocity Formula The velocity of the person with respect to the ground (V_person/ground) can be calculated using the formula: \[ V_{person/ground} = V_{person/train} + V_{train/ground} \] Where: - V_person/train = 1 m/s (the speed of the person relative to the train). - V_train/ground = 25 m/s (the speed of the train relative to the ground). ### Step 4: Substitute the Values Now, substituting the known values into the equation: \[ V_{person/ground} = 1 \, \text{m/s} + 25 \, \text{m/s} \] \[ V_{person/ground} = 26 \, \text{m/s} \] ### Step 5: Conclusion The speed of the person with respect to the ground is 26 m/s. ### Final Answer The speed of the person with respect to the ground is **26 m/s**. ---
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