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A man walks for some time ‘t’ withvelocity (𝑣) due east. Then he walks forsame time ‘t’ with velocity (v) due north.The average velocity of the man is:

Question

A man walks for some time ‘t’ with velocity v v due east. Then he walks for same time ‘t’ with velocity v v due north.

The average velocity of the man is:

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Solution

The average velocity is calculated by dividing the total displacement by the total time.

Step 1: Calculate the total displacement The man walks in two perpendicular directions (east and north) for the same amount of time with the same velocity. This forms a right-angled triangle where the two sides are of equal length (vt). The displacement is the hypotenuse of this triangle. Using Pythagoras' theorem, the displacement is √[(vt)² + (vt)²] = √[2(vt)²] = vt√2.

Step 2: Calculate the total time The total time is the sum of the time spent walking east and the time spent walking north, which is 2t.

Step 3: Calculate the average velocity The average velocity is the total displacement divided by the total time, which is (vt√2) / 2t = v√2 / 2.

So, the average velocity of the man is v√2 / 2.

This problem has been solved

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