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 ms^(-1)`
`v_(B)=0=v_(A)-g t_(AB)`
`therefore t_(AB)=(v_(A))/(g)=(20)/(10)=2s`
`therefore 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)/(2)g t_(BC)^(2)`
`therefore 70=(1)/(2)(10)t_(BC)^(2)`
`therefore t_(BC)= sqrt(14)=3.7 s`
`therefore t_(OABC)=7+3.7=10.7s`
Also `s_(OA)` = area under v - t graph between OA
`=(1)/(2)(5) (20)=50 m`