- #1

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So do I use [itex] \overline v_{x} = \frac{1}{2}(v_{x}_{0} + v_{x})t [/itex] or [itex] v_{x} = \frac{1}{2}(0 + \frac{47}{1.35})(1.35)[/itex] and do the same for the second part, except with different time and final speed?

Thanks

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- Thread starter courtrigrad
- Start date

- #1

- 1,235

- 1

So do I use [itex] \overline v_{x} = \frac{1}{2}(v_{x}_{0} + v_{x})t [/itex] or [itex] v_{x} = \frac{1}{2}(0 + \frac{47}{1.35})(1.35)[/itex] and do the same for the second part, except with different time and final speed?

Thanks

- #2

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any ideas?

- #3

lightgrav

Homework Helper

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the 47[m]/1.35

not the velocity at the END of interval 1.

While it might be reasonable to expect that

v_end = 2 * v_average .

The important point is that average velocity is

_defined_ to be (location change) / (time duration) !

you know the location change for time interval #2.

You also know the displacement for the two intervals

takean as one. Why are you trying to be fancy?

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