A rocket is accelerating straight up from the surface. At 1.35s after lift, the rocket clears top of its launch platform, 63m above the ground. After additional 4.45s its 1 km above ground. What's the magnitude of average velocity for the 4.45s part?
what's the magnitude of average velocity for the first 5.80 s of its flight?
what's the magnitude of average velocity for the first 5.80 s of its flight?
1 Answer
Jun 25, 2018
4.45s part:
5.80s part:
Explanation:
Notice that velocity is always in some unit of distance divided by some unit of time.
Average velocity = (total displacement)/(total time)
(Displacement has unit of distance.)
So, to solve the 4.45s part,
To solve the 5.80 s part,
I hope this helps,
Steve