In Q123, the net resultant ground speed of swimmer crossing along shortest path is:

Type: MCQ • Marks: 1

A. 2 km/h

B. 34 km/h

C. 8 km/h

D. 4 km/h

Explanation

Net velocity v_net = √(v² - u²) = √(5² - 3²) = √(25 - 9) = √16 = 4 km/h.

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UP TGT Mathematics

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