A rocket I fired upwards vertically with a net acceleration of `4m//s^(2)` and initial velocity zero.After 5 seconds its fuel is finsihed and it deccelerated 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:1To find velocity of rocket after 5 seconds.
`V_(A)=0+at_(OA)=(4)(5)=20ms^(-1)`
StageII:To find further time of ascent after 5 seconds .
0=2 `"gt"_(AB)" "therefore t_(AB)=(20)/(10)=2seconds`
Here,the total vertical displacement of stage (i) and
Stage (ii) is=area of OAB `=(1)/(2)(7)(20)=70 m.`