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))

A rocket is fired upwards vertically with a net acceleration of 4m//s^

1.	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))
Answer» Solution :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`