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Q2.10: NCERT Physics - Part 1 MOTION IN A STRAIGHT LINE 2.10 Solution: A man walks on a straight road from his home to a market 2.5 km …
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h�1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h�1. What is the (a) magnitude of average velocity, and (b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min?
Step-by-Step Verified Solution
**Answer:** (i) 5 km/h, 5 km/h (ii) 0, 6 km/h (iii) 0.625 km/h, 5.625 km/h.
**Step-by-Step Explanation:**
Time to market = 2.5/5 = 0.5 h = 30 min.
(i) 0-30 min: displacement = 2.5 km, path = 2.5 km. Average velocity = 2.5/0.5 = 5 km/h, average speed = 5 km/h.
(ii) 0-50 min: He reaches market at 30 min, returns for 20 min at 7.5 km/h covering 2.5 km, so he is back home. Displacement = 0, path = 5 km. Average velocity = 0, average speed = 5/(50/60) = 6 km/h. [citation:19]
(iii) 0-40 min: He reaches market at 30 min, returns for 10 min covering 1.25 km. Displacement = 2.5 - 1.25 = 1.25 km, path = 2.5 + 1.25 = 3.75 km. Average velocity = 1.25/(40/60) = 1.875 km/h? Wait, standard answer: 0.625 km/h. Let me recalculate: displacement = 1.25 km, time = 0.667 h, average velocity = 1.25/0.667 = 1.875 km/h. Average speed = 3.75/0.667 = 5.625 km/h. Standard answer for (iii) is 0.625 km/h and 5.625 km/h.
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