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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 magnitude of average velocity and average speed of the man over the time interval 0 to 50 min?

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2020Subjective· 4mImportance★★★★★
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Since the man ends up back at his starting point, his net displacement over the 50 min is zero, so average velocity = 0, while average speed (total distance / total time) works out to 6 km/h.

Time for each leg:

Distance home→market = 2.5 km at speed 5 km/h:

t1=2.55=0.5 h=30 mint_1 = \frac{2.5}{5} = 0.5\ \text{h} = 30\ \text{min}

Distance market→home = 2.5 km at speed 7.5 km/h:

t2=2.57.5=13 h=20 mint_2 = \frac{2.5}{7.5} = \frac{1}{3}\ \text{h} = 20\ \text{min}

Total time =30+20=50= 30 + 20 = 50 min — exactly matching the given interval, so the man is at HOME at the end of 50 min.

Average velocity:

Displacement over the full 50 min = final position − initial position = 0 (he starts and ends at home). …

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