1d Kinematics, missing something.

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Kildars
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A person who walks for exercise produces the position-time graph given with this problem.

Position-time graph of a walker: position x in km from 0 to +1.25 against time t in h from 0 to 1.00, with a line rising to +1.00 km by 0.20 h, flat until 0.40 h, falling to +0.25 km at 0.60 h, then rising to +1.25 km at 1.00 h, segments labeled A, B, C and D.


(b) Calculate the average velocity for each segment to verify your answers to part (a).
A
m/s
B
m/s
C
m/s
D
m/s

B is obviously 0

For A I did .20 of an hour is 300 seconds.
And 1km = 1000m

So 1000/300 =3.33 and that is wrong..

What am I doing wrong here I used:
[tex]v_{ave} = \Delta x / \Delta t[/tex]
 
The expression for average velocity is correct, but you calculated the time wrong. 1 hour equals 3600 seconds. So, 0.2 hours equals 720 seconds.
 
Ah, thank you.

I corrected my mistake and fixed my answers and they're correct.