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Physics · MOTION IN A STRAIGHT LINE
हिन्दीA man walks on a straight road from his home to a market \( 2.5\text{ km} \) away with a speed of \( 5\text{ km h}^{-1} \). Finding the market closed, he instantly turns around and walks back home with a speed of \( 7.5\text{ km h}^{-1} \). What are the magnitude of average velocity and average speed of the man over the time interval from \( 0 \) to \( 40\text{ min} \)?
Correct answer · Option 2
\( 1.875\text{ km h}^{-1} \) and \( 5.625\text{ km h}^{-1} \)
Explanation
Time to market = 30 min. In next 10 min (1/6 h), return distance = \( 7.5 \times \frac{1}{6} = 1.25\text{ km} \). Position = \( 1.25\text{ km} \) from home. Average velocity = \( \frac{1.25\text{ km}}{(40/60)\text{ h}} = 1.875\text{ km/h} \). Total distance = \( 2.5 + 1.25 = 3.75\text{ km} \). Average speed = \( \frac{3.75}{2/3} = 5.625\text{ km/h} \).
