What is the average velocity?

Question

An athlete swims the length of a 50.0-m pool in 29.0 s and makes the return trip to the starting position in 24.0 s. Take the initial direction of motion to be the positive direction. Determine her average velocities in the round trip in m/s.

Answer 1

Velocity = displacement / time.

Using the starting point as reference point,

1) for the outward trip:

Dispcement = +50m

Time = 29.0s

Velocity = +1.724m/s

Answer the velocity was 1.724 m/s in the direction of the swimmer.

2) For the return trip:

Displacement = 0

Time = 24.0s

Velocity = 0

3) Speed = 4*10^-6m/s

Distance = 9.6*10-²m

time =distance/speed

time = 9.6*10-²/4*10^-6

Time = 24,000 seconds

Time = 24000/3600 = 6.67 hours  

Answer 2

average velocity= displacement/time

50/29=1.72 m/s

50/24=2.08 m/s

Answer 3

100.0m/(29.0 + 24.0)s = 1.89m/s

9.6 cm is 9.6 x 10^-2 m

4.0 micrometers/s = 4.0 x 10^-6m/s

9.6 x 10^-2 / 4.0 x 10^-6 m/s = 2.4 x 10^4 s, or 24000 seconds, which is 400 minutes, or 6 hours, 40 minutes. Expressed in the two sig figs of the problem, 6.7 hours.

Answer 4

Velocity is a vector quantity which is displacement/time in a direction. Technically, her round trip velocity =0, because she returned to her starting point, and therefore displacement =0.

For each leg of the swim, her average velocity was:

+50/29 = +1.72 m/s

-50/24 = - 2.08 m/s

9.6e-2 m * 1 s/4.0e-6 m = 2.4e4 s

Answer 5

50m * 2 trips = 100m

total time= 53secons

100m/53seconds= 1.887m/s