A rocket is fired upwards vertically with a net acceleration of `4"m/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 = 10 m//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)`=70 m

Stage-iii : If `t_(BC)` is time of descent then
`70=1/2 (10)t_(BC)^2`
`therefore t_(BC)=sqrt14=3.7 s`