A small 8.00-kg rocket burns fuel that exerts a time-varying upward force on the rocket (assume constant mass) as the rocket moves upward from the launch pad. This force obeys the equation $\mathbf{F}\mathbf{=}\mathbf{A}\mathbf{+}{\mathbf{Bt}}^{\mathbf{2}}$ . Measurements show that at t= 0, the force is 100.0 N, and at the end of the first 2.00 s, it is 150.0 N.

(a) Find the constants Aand B, including their SI units. (b) Find the netforce on this rocket and its acceleration (i) the instant after the fuel ignites and (ii) 3.00 s after the fuel ignites. (c) Suppose that you were using this rocket in outer space, far from all gravity. What would its acceleration be 3.00 s after fuel ignition?

The mass of the rocket is

$\text{m}=8\text{kg}$

The time-dependent force from fuel ignition is given by the relation

$\text{F}=\text{A}+{\text{Bt}}^{\text{2}}$ (1)

The force at $\mathrm{t}=0\mathrm{s}$is

$\text{F}=100\text{N}$

The force at $\mathrm{t}=2\mathrm{s}$is

$\text{F}=150\text{N}$

The acceleration due to gravity is

$\text{g}=9.8\mathrm{m}/{\mathrm{s}}^2$

The downward gravitational force on a body of mass $m$ is

${\mathbf{F}}_{\mathbf{g}}\mathbf{=}\mathbf{mg}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(2\right)$

The second law of motion relates the net force $F$, mass $m$ and acceleration $a$as

$\mathbf{F}\mathbf{=}\mathbf{ma}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(3\right)$

Replace $\mathrm{t}=0\mathrm{s}$in equation (1) and get

$$\begin{gathered} 100\text{N}=\text{A}+\text{B}\times {\left(0\text{s}\right)}^2 \\ \text{A}=100\text{N} \end{gathered}$$

Replace, $\text{t}=2\text{s}$and value of $\mathrm{A}$in equation (1) and get

$$\begin{gathered} 150\text{N}=100\text{N}+\text{B}\times {\left(2\text{s}\right)}^2 \\ \text{B}=\frac{50\text{N}}{4{\text{s}}^{\text{2}}} \\ \text{B}=12.5·\left(1{\text{s}}^{-2}\right)·\left(1\text{N}\times \frac{{\displaystyle 1\text{kg}·\mathrm{m}/{\mathrm{s}}^2}}{1\mathrm{N}}\right) \\ \text{B}=12.5\text{kg}·\mathrm{m}/{\mathrm{s}}^4 \end{gathered}$$

Thus, A is 100 N and B is$12.5\text{kg}·\mathrm{m}/{\mathrm{s}}^4$.

Just after fuel ignition, two forces are acting on the body, the upward force from fuel ignition and the downward force from gravity. The net force is thus

$\begin{array}{rcl}{\mathrm{F}}_{\mathrm{n}}& =& \mathrm{F}\left(\mathrm{t}=0\right)-{\mathrm{F}}_{\mathrm{g}}\\ & =& 100\text{N}-\mathrm{mg}\\ & =& 100\text{N}-8\text{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\\ & =& 21.6\text{N}\end{array}$

From equation (3), the acceleration of the rocket is

$\begin{array}{rcl}\mathrm{a}& =& \frac{\mathrm{F}}{\mathrm{m}}\\ & =& \frac{21.6\text{N}}{8\text{kg}}\\ & =& 2.7·\left(1{\text{kg}}^{-1}\right)·\left(1\text{N}\times \frac{1\text{kg}·\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right)\\ & =& 2.7\mathrm{m}/{\mathrm{s}}^2\end{array}$

The net force is thus

$\begin{array}{rcl}{\mathrm{F}}_{\mathrm{n}}& =& \text{F}\left(\text{t}=\text{3}\right)-{\text{F}}_{\text{g}}\\ & =& 212.5\text{N}-\text{mg}\\ & =& 212.5\text{N}-8\text{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\\ & =& 134.1\text{N}\end{array}$

From equation (3), the acceleration of the rocket is

$\begin{array}{rcl}\mathrm{a}& =& \frac{\text{F}}{\text{m}}\\ & =& \frac{134.1\text{N}}{8\text{kg}}\\ & =& 16.76·\left(1{\text{kg}}^{-1}\right)·\left(1\text{N}\times \frac{1\text{kg}·{\text{m/s}}^2}{1\mathrm{N}}\right)\\ & =& 16.76{\text{m/s}}^2\end{array}$

Thus, the net force is 134.1 N and the acceleration is.16.76 m/s².

In space there is no gravity, thus the only force acting on the rocket is the force from fuel ignition.

The net force after 3 s is thus

$\begin{array}{rcl}{\mathrm{F}}_{\mathrm{n}}& =& \text{F}\left(\text{t}=3\right)\\ & & \text{=}212.5\text{N}\end{array}$

From equation (3), the acceleration of the rocket is

role="math" localid="1659441815075" $\begin{array}{rcl}\mathrm{a}& =& \frac{\text{F}}{\text{m}}\\ & =& \frac{212.5\text{N}}{8\text{kg}}\\ & =& 26.56·\left(1{\text{kg}}^{-1}\right)·\left(1\text{N}\times \frac{1\text{kg}·{\text{m/s}}^2}{1\mathrm{N}}\right)\\ & =& 26.56{\text{m/s}}^2\end{array}$

Thus, the acceleration is26.56 m/s².