A bike travels at 3 m/s over 2km. It then speeds up and travels at a higher velocity for another 2 km.
If the avg velocity for the entire 4 km trip is 4 m/s, find the avg velocity for the last 2 km. (Not 5 m/s)
How can I solve this without the time?
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3 responses so far ↓
1 gile // Sep 9, 2008
How can I solve this without the time?
You must find the time from the given information. After that you can evaluate speed.
2 kirchwey // Sep 9, 2008
You have to involve the time, either directly or indirectly.
Directly:
t1 = 2000 m / 3 m/s = 667 s
t(total) = 4000 m / 4 m/s = 1000 s
t2 = t(total)-t1 = 333 s
v2(ave) = 2000 m / 333 s = 6 m/s
Indirectly, we can do this without referring to t(anything) but it gets pretty cumbersome and harder to understand.
v2(ave) = (4000-2000) / (4000/4 - 2000/3) = 2000 / (1000 - 667) = 2000/333 = 6 m/s
3 intc_escapee // Sep 9, 2008
2000 = 3 t1
2000 = v t2
4000 = 4 (t1+ t2)
4000 = 4 (2000/3 + 2000/v)
1 = (2/3 + 2/v)
1/3 = 2/v
v = 6 m/s
Answer: v = 6 m/s
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