Physics: Arocket carrying a satellite is accelerating straight up from the earth s surface. at 1.15 s after liftoff, the rocket clears the top of its launch platform, 70 m above the ground. after an additional 4.70 s, it is 1.15 km above the ground. part a calculate the magnitude of the average velocity of the rocket for the 4.70 s part of its flight. express your answer in meters per second. part b calculate the magnitude of the average velocity of the rocket the first 5.85 s of its flight. express your

Arocket carrying a satellite is accelerating straight

chevalieriiim006

04.01.2022

a) v=230 m/s

b) v=196.5 m/s

Explanation:

a) The formula for average velocity is

$v=\frac{y_{2}-y_{1} }{t_{2}-t_{1} }$

For the first Δt=4.7s

$v=\frac{(1150-70)m}{(4.7)s} =230 m/s$

b) For the secont Δt=5.85s we know that the displacement is 1150m. So, the average velocity is:

$v=\frac{(1150)m}{(5.85)s}=196.5m/s$

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