A rocket is fired vertically upwards with a net acceleration of `4 m//s^2` and initial velocity zero. After `5s` its fuel is finished and it decelerates with g. At the highest point its velocity becomes zero. Then, it accelerates downwards with acceleration g and return back to ground. Plot velocity-time and displacement -time graphs for the complete journey. Take `g=10m//s^2.`

Explanation

In the graphs, `v_(A)=at_(OA)=(4)(5)=20 m//s " "v_(B)=0=v_(A)-g t_(AB)`
`:. t_(AB)=v_(A)/g=20/10=2s" " :. t_(OAB)=(5+2)s=7s`
Now, `s_(OAB)=` area under v-t graph between 0 to 7 s`=.^(1)//_(2)(7)(20)=70 m`
Now, `s_(OAB)=s_(BC)=1/2g t_(BC)^(2) " " :. 70=1/2(10)t_(BC)^(2)`
`:. t_(BC)=sqrt(14)=3.7 s " ":. t_(OABC)=7+3.7=10.7s`
Also `s_(OA)=` area under v-t graph between `OA=1/2(5)(20)=50 m`