As part of an exercise program, a woman walks south at a speed of 2.00 m/s for 60.0 minutes. she then turns around and walks north a distance 3000 m in 25.0 minutes (a) what is the woman's average velocity during her entire motion? a) 0.824 m/s south b) 1.93 m/s south c) 2.00 m/s south d) 1.79 m/s south e) 800 m/s south

Option A

Solution:

As per the question:

The distance covered by the woman in the North direction, d = 3000 m

Time taken to travel in North direction, t = 25.0 min = 1500 s

Velocity of woman in the south direction, v = 2.00 m/s

Time taken in the south direction, t' = 60.0 min = 3600 s

Now,

The distance covered in the south direction, d' = vt' =

Now, the total displacement is given by:

D = d' - d = 7200 - 3000 = 4200 m in South

(a) Average velocity of the woman in the whole journey is given by:

≈ 0.824 m/s South

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

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