Average Velocity

  • #1
courtrigrad
1,236
1
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 [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
 

Answers and Replies

  • #2
courtrigrad
1,236
1
any ideas?
 
  • #3
lightgrav
Homework Helper
1,248
30
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?
 

Suggested for: Average Velocity

  • Last Post
Replies
3
Views
1K
Replies
9
Views
570
Replies
1
Views
599
Replies
2
Views
369
Replies
3
Views
702
Replies
9
Views
320
Replies
16
Views
812
Replies
5
Views
595
  • Last Post
Replies
7
Views
534
Replies
23
Views
819
Top