Q40E - University Physics with Modern Physics

Question

A faulty model rocket moves in the x-y-plane (the positive y-direction is vertically upward). The rocket’s acceleration has components ${\mathit{a}}_{\mathbf{x}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{\alpha }{\mathit{t}}^{\mathbf{2}}$ and ${\mathit{a}}_{\mathbf{y}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{\beta }\mathbf{-}\mathit{\gamma }\mathit{t}$, where $\mathit{\alpha }\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{50}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{4}}$, and $\mathit{\gamma }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{40}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{3}}$. At t = 0 the rocket is at the origin and has velocity ${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}}_{\mathbf{0}}\mathbf{=}{\mathit{v}}_{\mathbf{0}\mathbf{x}}\hat{\mathbf{i}}\mathbf{+}{\mathit{v}}_{\mathbf{0}\mathbf{y}}\hat{\mathbf{j}}$ with v_0x = 1.0m/s and v_0y = 7.0m/s. (a) Calculate the velocity and position vectors as functions of time. (b) What is the maximum height reached by the rocket? (c) What is the horizontal displacement of the rocket when it returns to y = 0?

Step-by-Step Solution

Answer

1. The position of the rocket is $x={x_0}_{x}t+\frac{1}{12}\alpha t^4$ and $y=y_{0y}t+\frac{1}{2}\beta t^2-\frac{1}{6}\gamma t^3$, the velocity as a function time is $v_{0x}+\frac{1}{3}\alpha t^3+v_{0y}+\beta t-\gamma t^2$.
2. The maximum height of the rocket is 340.6 m
3. The horizontal displacement of the rocket is 3.85x10⁴ m

Step 1: Identification of given data

The given data can be listed below,

• The value of constant is, $\alpha =2.50m/s^4$
• The value of the constant for y-component is, $\beta =9.0m/s^2$
• The initial x-component of velocity of rocket is, $v_{0x}=1.0m/s$
• The initial y-component of velocity of rocket is, $v_{0y}=7.0m/s$

Step 2: Concept/Significance of position vector.

An object's position may be described using its position vector. It is essential to understand a body's location in order to accurately describe its motion.

Step 3: (a) Determination of the velocity and position vectors as functions of time.

The velocity of the rocket is given by integrating acceleration which is given as,

${\int }_{v_0}^{v}dv={\int }_{v_0}^{v}adt$

Substitute all the values in the above, 

The x-component of the velocity is calculated as,

$\begin{array}{rcl}{v}_x-v_{0x}& =& {\int }_0^t\alpha t^{2}dt\\ v_x& =& v_{0x}+\frac{1}{3}\alpha t^2\end{array}$

The y-componentsof the rocket is given by,

$\begin{array}{rcl}{\int }_{v_0}^{v}d{v}_y& =& {\int }_0^t\left(\beta -\gamma t\right)dt\\ v_y-v_{0y}& =& {\overline{)\left(\beta t-\frac{1}{2}\gamma t^2\right)}}_0^t\\ v_y& =& v_{0y}+\beta t-\gamma t^2\end{array}$

The velocity vector of the rocket is given by,

$\stackrel{\rightharpoonup }{v}\left(t\right)=v_{0x}+\frac{1}{3}\alpha t^3+v_{0y}+\beta t-\gamma t^2$

The x-component of the position is given by,

$\begin{array}{rcl}{v}_x& =& \frac{dx}{dt}\\ x-x_0& =& {\int }_0^t{v}_{x}dt\end{array}$

Substitute all the values in the above,

$\begin{array}{rcl}x-x_0& =& {\int }_0^t{v}_{0x}+\frac{1}{3}\alpha t^{3}dt\\ & =& v_{ox}t+\frac{1}{12}\alpha t^4\\ x& =& x_0+v_{0x}t+\frac{1}{12}\alpha t^4\end{array}$

The y-component of position is given by,

$$\begin{gathered} y-y_0={\int }_0^t\left(v_{0y}+\beta t-\frac{1}{2}\gamma t^2\right)dt \\ =v_{oy}t+\frac{1}{2}\beta t^2-\frac{1}{6}\gamma t^3 \\ y=y_0+v_{0y}t+\frac{1}{2}\beta t^2-\frac{1}{6}\gamma t^3 \end{gathered}$$

Thus, the position of the rocket is $\begin{array}{rcl}x& =& x_0+v_{0x}t+\frac{1}{12}\alpha t^4\end{array}$ and $y=y_0+v_{0y}t+\frac{1}{2}\beta t^2-\frac{1}{6}\gamma t^3$, the velocity as a function time is $v_{0x}+\frac{1}{3}\alpha t^3+v_{0y}+\beta t-\gamma t^2$.

Step 4: (b) Determination of the maximum height reached by the rocket

At the maximum height, the vertical component of velocity is zero. The heights of the rocket is given by,

$\begin{array}{rcl}{v}_0& =& v_{0y}+\beta t-\frac{1}{2}\gamma t^2\\ 0& =& v_{0y}+\beta t-\frac{1}{2}\gamma t^2\end{array}$

Substitute all the values in the above,

$\begin{array}{rcl}0& =& 7m/s+\left(9m/s^2\right)t-\frac{1}{2}\left(1.40m/s^2\right)t^2\\ t& =& 13.6 s\end{array}$

Substitute all the values in the equation for y-component of position is given by,

$\begin{array}{rcl}y& =& 7\times 13.6+\frac{9\times {\left(13.6\right)}^2}{2}-\frac{1.4\times {\left(13.6\right)}^3}{6}\\ & =& 340.6m\end{array}$

Thus, the maximum height of the rocket is 340.6 m

Step 4: (c) Determination of the horizontal displacement of the rocket when it returns

The time of the rocket by the equations given by,

$\begin{array}{rcl}{v}_{0y}t+\frac{1}{2}\beta t^2-\frac{1}{6}\gamma t^3& =& 0\\ \frac{1}{6}\gamma t^2-\frac{1}{2}\beta t-v_{0y}& =& 0\end{array}$

Substitute all the values in the above,

$\begin{array}{rcl}\left(1.40\right)\frac{1}{6}{t}^2-\frac{1}{2}\left(9.0\right)t-v_{0y}& =& 0\\ t& =& 20.73 s\end{array}$

Put the value of time in the equation for x-component of position, $\begin{array}{rcl}x& =& 1\times 20.73+\frac{2.5\times {\left(20.73\right)}^4}{12}\\ & =& 38500m\\ & =& 3.85\times {10}^{4}m\end{array}$

The graph for trajectory is shown below as,

From graph it is clear that the x and y components are not symmetric.

Thus, the horizontal displacement of the rocket is 3.85x10⁴ m