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?

Answer 1

0.824m/s

Step-by-step explanation:

To calculate the average velocity we have to find the total distance and the total time

We have to find the distance and time in each motions

FIRST MOTION

The values given are

Speed= 2m/s , t = 60minutes

The time has to be converted to seconds. 60×60 = 3600seconds

Distance= speed×time

= 2× 3600

= 7200m

In the first motion the distance is 7200m and the time is 3600seconds

SECOND MOTION

The values given are

Distance= 3000m

Time= 25 mins to seconds

= 25×60

= 1500 seconds

In the second motion distance is 3000m and the time is 1500 seconds

The total distance can be calculated by applying the formular ( d1-d2) since she moved in an opposite direction

Total distance= 7200-3000

= 4200m

The total time (t1+t2) = 1500+3600

= 5100 seconds

Therefore, average velocity is calculated by applying the formular

Total distance/ Total time

= 4200/5100

= 0.824m/s South

Hence the average velocity is 0.824m/s South.

Answer 2

The woman's average velocity during her entire motion is 2 m/s

Step-by-step explanation:

Given;

initial speed of the woman, u = 2.00 m/s

initial time taken, t₁ = 60 minutes = 3600 seconds

initial displacement of the woman, x₁ = ?

final displacement of the woman, x₂ = 3000 m north

final time taken , t₂ = 25.0 minutes = 1500 seconds

The woman's average velocity during her entire motion:

initial displacement of the woman, x₁ = u x t₁ = 2.00 m/s x 3600 seconds

= 7200 m South

`Average \ velocity = (\delta X)/(\delta t) = (X_1-X_2)/(t_1-t_2) \\\\V_(avg.) = (7200-3000)/(3600-1500) = (4200)/(2100) = 2 \ m/s`

Therefore, the woman's average velocity during her entire motion is 2 m/s