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winterhawk8 · Jun 30, 2009, 12:08 PM

How do we figure out distance travelled.
18.0m(W) in a time of 6.00s and stopped. After stopping, it moved 12.0m(9E) in 10.0s

Perito · Jun 30, 2009, 02:02 PM

How do we figure out distance traveled?

moved 18.0m(W) in a time of 6.00s and stop
After stopping, moved 12.0m(9E) in 10.0s

Distance = Rate x Time.

You are given two distances and two times. You can solve the above equation for rate and get that:

$Rate = \frac {Distance}{Time}$

$\frac {18\,m}{6.00\,s}$

$\frac {12\,m}{10.0\,s}$

I'm afraid I don't understand your question, "How do we figure out distance traveled?" The distances are given, 18.0 m and 12.0 m. The sum is 30.0 m and that's the distance you traveled. If you are after an average velocity (average rate), take the total distance (18 m + 12 m = 30.0 m) and divide by the total time (6.0 s + 10.0 s = 16.0 s). The average velocity is, therefore

$Average\, velocity = \frac {30.0\,m}{16.0\,s} = 1.875\,\frac {m}{s}$

Unknown008 · Jul 1, 2009, 07:00 AM

Do you mean that the velocities are:
18 m/s for 6 s towards the west, and
12 m/s for 10 s towards the east?

Then, the distance travelled is 18+12 = 30 m
The displacement is 18-12 = 6 m to the west (taking the western direction as positive)
The average velocity will then be 6/16 = 0.375 m/s