Question

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.`

Answer 1

In the graphs, `v_A = at_(OA) = (4)(5) = 20 m//s`,
`" "v_B = 0 =v_A - "gt" _(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) "gt"^(2)_(BC)" " therefore 70 = (1)/(2)(10) t^(2)_(BC) `
`therefore " "t_(BC) = sqrt(14)= 3.7 s" " therefore t_(OAB) = 7+ 3.7 = 10.7 s`
Also `s_(OA)` = area under v-t graph between `OA = (1)/(2) (5)(20) = 50` m