A rocket is fired upwards vertically with a net acceleration of `4m//s^(2)` and initial velocity zero. After 5 seconds its fuel is finished and it decelerates with g. At the highest point its velocity becomes zero. There after it accelerates downwards with acceleration g and return back to ground.
i) Plot velocity - time graph for complete journey
ii) Displacement-time graph for the complete journey. (Take `g=10m//s^(2)`)

Explanation

Stage I : To find velocity of rocket after 5 seconds
`V_(A)=0+at_(OA)=(4)(5)=20ms^(-1)`
Stage II : To find further time of ascent after 5 seconds.
`0=20-g t_(AB)`
`therefore t_(AB)=(20)/(10)=2`seconds
Here, the total vertical displacement of stage (i) and stage (ii) is
= area of `OAB=(1)/(2)(7)(20)=70m`.

Stage-iii : If `t_(BC)` is time of descent then
`70=(1)/(2)(10)t_(BC)^(2)`
`therefore t_(BC)=sqrt(14)=3.7s`