Question

A two stage rocket is launched moving vertically with acceleration 5.0 m/s^2. After 10.0 s, the first stage of the rocket is ejected; the second stage is now accelerating at 8.0 m/s^2. What is the distance between the first and second stages 4.0 s after separation?

Answer 1

Answer:264 m

Step-by-step explanation:

Given

acceleration of rocket(
`a_1`)= 5 m/s^2[/tex]

velocity after 10 s

v=u+at

`v=0+5* 10`

v=50 m/s

after first stage rocket is ejected

acceleration of second stage=
`8 m/s^2`

distance between first and second part after 4 sec

`s=u_1t+(1)/(2)at^2`

here
`u_1=50 m/s`

`s=50* 4+(1)/(2)* 8* 4^2`

s=200+64=264 m