Answer:264 m
Explanation:
Given
acceleration of rocket([tex]a_1[/tex])= 5 m/s^2[/tex]
velocity after 10 s
v=u+at
[tex]v=0+5\times 10[/tex]
v=50 m/s
after first stage rocket is ejected
acceleration of second stage=[tex]8 m/s^2[/tex]
distance between first and second part after 4 sec
[tex]s=u_1t+\frac{1}{2}at^2 [/tex]
here [tex]u_1=50 m/s[/tex]
[tex]s=50\times 4+\frac{1}{2}\times 8\times 4^2[/tex]
s=200+64=264 m
