# Average Velocity

A rocket carrying a satelite is accelerating straight up from the Earth's surface. The rocket clears the top of its launch platform, 47 m above the ground, 1.35 seconds after lift off. After an additional 4.45 seconds it is 1.00 km above the ground. Calculate the magnitude of the average velocity of the rocket for (a) the 4.45 second part of the flight; (b) the first 5.80 second part of the flight.

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

Thanks

any ideas?

Homework Helper
No, you don't.

the 47[m]/1.35 is the average velocity of interval 1,
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?