Question

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

Answer 1

Answer:

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' = $2.00\times 3600 = 7200\ m$

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:

$v_{avg} = \frac{Total\ displacement}{Total\ time} = \frac{4200}{t + t'}$

$v_{avg} = \frac{4200}{1500 + 3600} = 0.8235\ m/s$ ≈ 0.824 m/s South