91Ó°ÊÓ

• The thrust produced by engines = 3.50×10⁷ N.

Apply Newton’s Second Law of motion as:

$\begin{array}{c}{F}_{net}=ma\\ T−mg=ma\end{array}$

Here, m is the mass of the rocket, a is the acceleration, T is the engine thrust, g is the acceleration due to gravity, and F_net is the net force exerted.

Substitute 2x10⁶ kg for m, 3.5x10⁷ N forT, and 9.8 m/s² for g in the above expression, and we get,

$\begin{array}{c}3.5×{10}^7\text{ N}−2×{10}^6\text{ k²µ}×{\text{9.8 m/²õ}}^2=2×{10}^6\text{ k²µ}×a\\ \left(3.5×{10}^7−1.96×{10}^7\right)\text{ k²µ}â‹\dots {\text{m/s}}^{\text{2}}=2×{10}^6\text{ k²µ}×a\\ a=\frac{1.54×{10}^7\text{ k²µ}â‹\dots {\text{m/s}}^{\text{2}}}{2×{10}^6\text{ k²µ}}\\ a=7.7{\text{ m/²õ}}^2\end{array}$

Hence, the initial acceleration of the rocket is 7.7 m/s².

Convert the speed from 120 km/h to m/s as:

$\begin{array}{c}120\text{ k³¾/³ó}=\text{120 }\frac{\text{km}}{\text{h}}×\frac{1000\text{ m}}{1\text{ k³¾}}×\frac{1\text{â€É¾}}{3600\text{ s}}\\ =33.33\text{ m/²õ}\end{array}$

Apply the equation of motion as:

$\mathit{v}\mathbf{=}\mathit{u}\mathbf{+}\mathit{a}\mathit{t}$

Here, v is the final speed, u is the initial speed, and t is the time taken.

Substitute 0 m/s for u, 7.7 m/s²for a, and 33.33 m/s for vin the above expression, and we get,

$\begin{array}{c}33.33\text{ m/²õ}=0+7.7{\text{ m/²õ}}^2×t\\ t=\frac{33.33\text{ m/²õ}}{7.7{\text{ m/²õ}}^2}\\ t=4.329\text{ s}\end{array}$

Hence, the time taken by the rocket is 4.329 s.

When the mass of the rocket decreases, the acceleration of the rocket increases for the same value of thrust. With the increased acceleration, it takes less time than 4.329 s.