Question

A diwali rocket moves vertically up with a constant acceleration $a_1$ = 20/3 $m{s}^{-2}$. After sometimes, its fuel gets exhausted and then it falls freely with an acceleration $a_2=10m{s}^{-2}$. If the maximum height attained by the diwali rocket is h, using graphical method, find its speed when the fuel is just exhausted. Assume h = 50 m.

Answer 1

Let the rocket acquire a speed v at the time when all its fuel gets exhausted. Referring to the v-t graph in Fig., the slope of OP gives the acceleration $a_1$.
The slope of OP gives the acceleration
$a_1=tan{\Theta }_1=\frac{v}{t_1}$ ...(i)
The slope of PQ gives the acceleration
$-\left(a_2\right)=tan{\Theta }_2=\frac{-v}{t_2}$ ...(ii)
From (i) and (ii), we have
$v=a_1{t}_2=a_2{t}_2$. ...(iii)
The area under v - t graph = Area of $\Delta$OPQ = (1/2) OQ.PM = s
Substituting OQ = $t_1+t_2$ and PM = v,
The total displacement s =(v/2)$(t_1+t_2$ ...(iv)
Substituting $t_1=\frac{v}{a_1}$ and $t_2=\frac{v}{a_2}$ from(iii) in (iv), we have
$s=\frac{v}{2}(\frac{v}{a_1}+\frac{v}{a_2})$
Finally substituting s = +h (as the area is positive), $a_1=\frac{20}{3}m{s}^{-2}$ and $a_2=10m{s}^{-2}$, we have
$v=\sqrt{\frac{2a_1{a}_{2}h}{a_1+a_2}}=\sqrt{\frac{2\times \left(20/3\right)\times 10\times 50}{\frac{20}{3}+10}}$
This yields v = 20 m/s.