Average Velocity of a Rocket Launch: Calculating Magnitude and Time

AI Thread Summary
To calculate the average velocity of a rocket during its ascent, the correct approach is to use the formula for average velocity defined as the change in location divided by the time duration. For the first interval of 1.35 seconds, the average velocity is determined by the height of 47 meters divided by the time. For the second interval of 4.45 seconds, the total height of 1.00 km minus the initial height can be used to find the average velocity for that segment. It is emphasized that average velocity should not be confused with final velocity at the end of an interval. Understanding these calculations is crucial for accurate analysis of rocket motion.
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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
 
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any ideas?
 
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?
 
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