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Q.A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 kmh^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 kmh^-1. What is the

(a) magnitude of average velocity and
(b) average speed of the man over the time interval 0 to 50 minutes ?
Andhra Pradesh BieapBIEAP Intermediate Board (1st Year) 2018Subjective· 4mImportance★★★★★
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Average velocity = 0 (he ends where he started); average speed = 6 km/h.

Step 1 — Find the times for each leg:

Going to market: distance = 2.5 km at 5 km/h, so time = 2.5/5 = 0.5 h = 30 minutes.

Returning home: distance = 2.5 km at 7.5 km/h, so time = 2.5/7.5 = 1/3 h = 20 minutes.

Total time = 30 + 20 = 50 minutes. So the interval 0 to 50 min covers the full round trip and he is back home at t = 50 min.

Step 2 — (a) Average velocity:

Average velocity = total displacement / total time.

Since he starts at home and ends at home, the displacement over 0 to 50 min is zero.

Therefore average velocity = 0 / (50 min) = 0.

Step 3 — (b) Average speed: …

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