A rocket is launched from the suface of the earth vertically upward from rest. It moves under acceleration `8 m//s^(2)`. After 15 s the fuel is finished, the rocket moves under gravity and ultimately strickes the ground. Assuming upward direction position and launching point as the origin, sketch the v-t, s-t and a-t graph.

Explanation

Here the motion is in two parts: first acceleration and then motion under gravity.

Acceleration motion
O to A: `h_(1)=0+1/2a_(1)t_(1)^(2)=1/2xx8xx(15)^(2)=900 m`
A to O through B
A: Origin
Displacement`=-900 m`
`S=ut-1/2 at^(2)`
`-900=v_(0)t_(2)-1/2 a^(2)t_(2)^(2)=120 t_(2)-5t_(2)^(2)`
`t_(2)^(2)-24t_(2)-180=0`
`t_(2)=(24+-sqrt((-24)^(2)-4(1)(-180)))/2`
`=(24+-36)/2`
`t_(2)=30 s`, the rocket will strike the ground after 30 s (after the fuel is finished)
`v^(')=u-g t_(2)`
`=v_(0)-g t_(2)`
`=120-10xx30=-180 m//s`
The velocity of the rocket will be zero after time `t_(3)`. So,
`0=120-10t_(3)impliest_(3)=12 s`
To obtain the maximum height attained by the rocket above A, we have
`0=(v_(0))^(2)-2xx10xxh_(2)impliesh_(2)=720 m` at time
`t^(')=12 s`
Total time of journey `=15+30=45 s`
v-t graph
`t=0, v=0`
`t=15 s, v=120 m//s`
`t=27 s, v=0`
`t=45 s, v=-180 m//s`
`t=0` to `t=15 s`, `v=8t` (straight line)
`t=15` to `t=45s`
`v=120-10(t-15)` (straight line) `[15 s lt = t lt = 45 s]`

s-t graph
`t=0, s=0`
`t=15 s, s=900 m`
`t=27 s, s, s=900+720=1620 m`
`t=45 s, s=0`
`t=0` to `t=15 s`, `s=1/2 a_(1)t^(2) implies s=4t^(2)`
Shape: `y=4x^(2)` (parabola, open upward)
`t=15 to 45 s`
`s=900+v_(0)(t-15)-1/2a_(2)(t-15)^(2)`
`=900+120(t-15)-5(t-15)^(2)`
Shape: `y=900+120x-5x^(2)` (parabola, open downward)