Question

A rocket 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 answer in meters per second.

Answer 1

Answer:

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:

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