Q 22E - University Physics with Modern Physics

Question

Question: A 2540 kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force such that its vertical velocity as a function of time is given by $\mathit{v}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{A}\mathit{t}\mathbf{+}\mathit{B}{\mathit{t}}^{\mathbf{2}}$, where A and B are constants and time is measured from the instant the fuel is ignited. The rocket has an upward acceleration of $\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{ }{\mathbf{\text{m/s}}}^{\mathbf{\text{2}}}$ at the instant of ignition and,1.00 s later, an upward velocity of $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\text{m/s}}$. (a) Determine A and B, including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newton and as a multiple of the rocket’s weight. (d) What was the initial thrust due to the fuel?

Step-by-Step Solution

Answer

a) The value of A and B is, $1.5 {\text{m/s}}^{\text{2}} \text{and} 0.5 {\text{m/s}}^{\text{3}}$.

b) The acceleration of the rocket at $4.00 \text{s}$ is, $5.5 {\text{m/s}}^{\text{2}}$.

c) Thrust force on rocket is, $3.886\times {10}^4 \text{N}$.

d) The initial thrust force due to fuel is,  $2.87\times {10}^4 \text{N}$.

Step 1: Identification of given data

The given data listed below as,

• The mass of rocket is,  $m=2.54\times {10}^3 \text{kg}$.
• The vertical velocity function is, $v\left(t\right)=At+B{t}^2$.
• The acceleration of rocket at the instant of ignition is, $a=1.5 {\text{m/s}}^{\text{2}}$.
• The upward velocity of rocket is,  $v=2 \text{m/s}$.

Step 2: Significance of thrust force.

Thrust is the force produced by the combustion of pressurized gasses in the engine and then the release of these gasses. The thrust produced helps the rocket attain escape velocity to move out of the atmosphere of the earth.

Step 3: Determination of A and B.

The expression for the vertical velocity function can be expressed as,

 $v\left(t\right)=At+B{t}^2$

 

Here A and B are the constants, and t is the time.

 

Substitute $1.00 \text{s}$ for t in the above equation, we get-

$\begin{array}{rcl}v& =& A (1.00 \text{s})+B (1.00 \text{s}{)}^2\\ & =& 2 \text{m/s}\end{array}$

 

Hence vertical velocity is, 2m/s.

 

The relation for the acceleration can be expressed as,

$a\left(t\right)=\frac{dv}{dt}$

 

Substitute $At+B{t}^2$ for v in and differentiate with respect to t in the above equation.

 $$\begin{gathered} a\left(t\right)=\frac{d }{dt}(At+B{t}^2) \\ a\left(t\right)=A+2Bt \end{gathered}$$

 

Substitute 0 s for t in order to solve for initial condition, and $1.5 {\text{m/s}}^{\text{2}}$ for a in the above equation.

 $$\begin{gathered} 1.5 {\text{m/s}}^{\text{2}}=A+0 \\ A=1.5 {\text{m/s}}^{\text{2}} \end{gathered}$$

 

Hence, the required value of A is, $1.5 {\text{m/s}}^{\text{2}}$.

 

The expression for the vertical velocity can be expressed as,

$v\left(t\right)=At+B{t}^2$

 

Substitute 1.00 s for t, $1.5 {\text{m/s}}^{\text{2}}$ for A , and 2m/s for v in the above equation.

$$\begin{gathered} 2 \text{m/s}=(1.5 {\text{m/s}}^{\text{2}}) (1.00 \text{s})+B (1.00 \text{s}) \\ B=0.5 {\text{m/s}}^{\text{3}} \end{gathered}$$

 

Hence, required value of B is, $0.5 {\text{m/s}}^{\text{3}}$.

Step 4: Determination of acceleration at t = 4 . 00   s .

The expression for the acceleration can be expressed as,

 $$\begin{gathered} a\left(t\right)=\frac{dv}{dt} \\ a=A+2Bt \end{gathered}$$

  

Substitute 4.00 s for t, $1.5 {\text{m/s}}^{\text{2}}$  for A, and $\text{0.5} {\text{m/s}}^3$ for B in the above equation.

 $\begin{array}{rcl}a& =& 1.5 {\text{m/s}}^{\text{2}}+2 (0.5 {\text{m/s}}^{\text{3}}) (4.00 \text{s})\\ & =& 5.5 {\text{m/s}}^{\text{2}}\end{array}$

 

Hence, required acceleration is, $5.5 {\text{m/s}}^{\text{2}}$.

Step 5: Determination of thrust force.

The expression for the thrust force acting on rocket can be expressed as,

 $\begin{array}{rcl}F& =& m a+W\\ & =& m a+m g\end{array}$

Here m  is the mass, and a is the acceleration, g is the acceleration due to gravity.

 

Substitute $2.54\times {10}^3 \text{kg}$ for m, $5.5 {\text{m/s}}^{\text{2}}$ for a, and $9.8 {\text{m/s}}^{\text{2}}$ for g in the above equation.

 $\begin{array}{rcl}F& =& (2.54\times {10}^3 \text{kg)} \text{(5.5} {\text{m/s}}^2)+(2.54\times {10}^3 \text{kg)} \text{(9.8} {\text{m/s}}^2)\\ & =& 3.886\times {10}^4 \text{N}\end{array}$

  

Hence, required thrust force is,  $3.886\times {10}^4 \text{N}$.

Step 6: Determination of initial thrust due to fuel.

The expression for the thrust force due to fuel can be expressed as,

$\begin{array}{rcl}F& =& m a_{rocket}+W\\ F& =& m a_{rocket}+m g\\ & =& m (a_{rocket}+g)\end{array}$

 

Substitute $2.54\times {10}^3 \text{kg}$ for m, $1.5 {\text{m/s}}^{\text{2}}$ for $a_{rocket}$, and $9.8 {\text{m/s}}^{\text{2}}$ for g in the above equation.

$\begin{array}{rcl}F& =& 2.54\times {10}^3 \text{kg} \text{(1.5} {\text{m/s}}^2+9.8 {\text{m/s}}^{\text{2}})\\ & =& 2.87\times {10}^4 \text{N}\end{array}$  

Hence, required thrust force due to fuel is, $2.87\times {10}^4 \text{N}$.