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

a)
`v=230 m/s`

b)
`v=196.5 m/s`

Step-by-step explanation:

a) The formula for average velocity is

`v=(y_(2)-y_(1)  )/(t_(2)-t_(1)  )`

For the first Δt=4.7s

`v=((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=((1150)m)/((5.85)s)=196.5m/s`