Q39P - University Physics with Modern Physics

Question

A rocket is fired at an angle from the top of a tower of height ${\mathit{h}}_{\mathbf{0}}\mathbf{=}\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{m}}$. Because of the design of the engines, its position coordinates are of the form $\mathit{x}\left(t\right)\mathbf{=}\mathit{A}\mathbf{+}\mathit{B}{\mathit{t}}^{\mathbf{2}}$ and $\mathit{y}\left(t\right)\mathbf{=}\mathit{C}\mathbf{+}\mathit{D}{\mathit{t}}^{\mathbf{3}}$y1t2 = C + Dt3, where A, B, C, and D are constants. The acceleration of the rocket 1.00 s after firing is $\overrightarrow{\mathbf{a}}\mathbf{=}\left(4.0\hat{i}+3.0\hat{j}\right)\mathbf{ }{\mathbf{\text{m/s}}}^{\mathbf{2}}$. Take the origin of coordinates to be at the base of the tower. (a) Find the constants A, B, C, and D, including their SI units. (b) At the instant after the rocket is fired, what are its acceleration vector and its velocity? (c) What are the x- and y-components of the rocket’s velocity 10.0 s after it is fired, and how fast is it moving? (d) What is the position vector of the rocket 10.0 s after it is fired?

Step-by-Step Solution

Answer

1. The values for constants and their SI units are $A=0, B=2 {\text{m/s}}^2\text{,} \text{C}=50 \text{m} \text{and} D=0.5 {\text{m/s}}^3$
2. The velocity and acceleration of the rocket after being fired are,$0$ and $4 \text{m/}{\text{s}}^2\hat{i}$ respectively.
3. The x and y components of velocity are $40 \text{m/s}$ and $150 \text{m/s}$.
4. The position vector of the rocket after firing is $\left(200\hat{i}+550\hat{j}\right) \text{m}$.

Step 1: Identification of given data

The given data can be listed below,

• The rocket is fired at a height at,$h_0=500 \text{m}$.
• The acceleration of rocket is,$\overrightarrow{a}=4\hat{i}+3\hat{j} {\text{m/s}}^2$.

Step 2: Concept/Significance of wind speed

The difference in pressure over a given distance directly relates to the wind's speed. The stronger the wind, the bigger the difference or shorter the distance.

Step 3: (a) Determination of the constants and their SI unit.

The position of the rocket is given by,

$r\left(t\right)=\left(x\left(t\right),y\left(t\right)\right)$

Here,$x\left(t\right)$ is the x-coordinate of rocket, and $y\left(t\right)$ is the y-coordinate the rocket.

The velocity of the vector is given by,

$v\left(t\right)=\frac{d\overrightarrow{r}}{dt}$

Substitute all the values in the above,

$$\begin{gathered} v\left(t\right)=\frac{d\left(A+B{t}^2, C+D{t}^3\right)}{dt} \\ =\left(2Bt, 3D{t}^3\right) \end{gathered}$$

The acceleration of the rocket is given by,

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

Substitute all the values in the above,

$$\begin{gathered} a\left(t\right)=\frac{d\left(2Bt+3D{t}^2\right)}{dt} \\ =2Bt+6Dt \end{gathered}$$

Initially the position of rocket is $h_0$ and acceleration at 1 s is (4, 3).

Substitute the values in above,

$A = 0$

$$\begin{gathered} \overrightarrow{r}\left(t\right)=\left(A+B{t}^2, C+D{t}^3\right) \\ \left(0,50 \text{m}\right)=\left(0+B\left(0\right), C+D\left(0\right)\right) \\ C=50 \text{m} \end{gathered}$$

From acceleration, the value of B is given by,

$$\begin{gathered} 2B=4 {\text{m/s}}^2 \\ B=2 {\text{m/s}}^2 \end{gathered}$$

From the value of D put values in the equation for acceleration,

$$\begin{gathered} 3 {\text{m/s}}^2=6D\left(1 \text{s}\right) \\ D=\frac{3}{6} {\text{m/s}}^2 \\ =0.5 {\text{m/s}}^3 \end{gathered}$$

Thus, the values for constants and their SI units are $A=0, B=2 {\text{m/s}}^2\text{,} \text{C}=50 \text{m} \text{and} D=0.5 {\text{m/s}}^3$

Step 4: (b) Determination of the acceleration vector and velocity of the rocket after firing.

The acceleration and velocity of the rocket is determined by substituting the value of constant in to the equation for acceleration and velocity.

The equation of velocity and acceleration is given by,

$$\begin{gathered} \overrightarrow{r}\left(t\right)=\left[\left(2 {\text{m/s}}^2\right)t^2\right]\hat{i}+\left[50 \text{m}+\left(0.5 {\text{m/s}}^3\right)t^3\right]\hat{j} \\ \overrightarrow{v}\left(t\right)=\left[\left(4 {\text{m/s}}^2\right)t\right]\hat{i}+\left[\left(1.5 {\text{m/s}}^3\right)t^2\right]\hat{j} \\ \overrightarrow{a}\left(t\right)=\left(4 {\text{m/s}}^2\right)\hat{i}+\left[\left(3 {\text{m/s}}^3\right)t\right]\hat{j} \end{gathered}$$

Substitute all the values in the above equations the initial velocity of rocket after being fired is,

$$\begin{gathered} \overrightarrow{v}\left(t\right)=\left[\left(4 {\text{m/s}}^2\right)\left(0\right)\right]\hat{i}+\left[\left(1.5 {\text{m/s}}^3\right)\left(0\right)\right]\hat{j} \\ =0 \end{gathered}$$

The initial acceleration of rocket after being fired is,

$$\begin{gathered} \overrightarrow{a}\left(t\right)=\left(4 {\text{m/s}}^2\right)\hat{i}+\left[\left(3 {\text{m/s}}^3\right)\left(0\right)\right]\hat{j} \\ =4 \text{m/}{\text{s}}^2\hat{i} \end{gathered}$$

Thus, the velocity and acceleration of the rocket after being fired are, $0$and $4 \text{m/}{\text{s}}^2\hat{i}$ respectively.

Step 5: (c) Determination of the x- and y-components of the rocket’s velocity 10.0 s after it is fired, and how fast is it moving

The x-component of the velocity of rocket is given by,

$v_x=\left(4 {\text{m/s}}^2\right)t$

Substitute the value of time in above,

$$\begin{gathered} v_x=\left(4 {\text{m/s}}^2\right)\left(10 \text{s}\right) \\ =40 \text{m/s} \end{gathered}$$

The y-component of velocity of the rocket is given by,

$v_y=\left(1.5 {\text{m/s}}^2\right)t^2$

Substitute the value of time in the above,

$$\begin{gathered} v_y=\left(1.5 {\text{m/s}}^2\right){\left(10 \text{s}\right)}^2 \\ =150 \text{m/s} \end{gathered}$$

Thus, the x and y components of velocity are $40 \text{m/s }$and $150 \text{m/s}$.

Step 5: (d) Determination of theposition vector of the rocket 10.0 s after it is fired

The position vector of the rocket after 10 s after being fired is given by,

$$\begin{gathered} r\left(t\right)=\left[\left(2 {\text{m/s}}^2\right)t^2\right]\hat{i}+\left[50 \text{m}+\left(0.5 {\text{m/s}}^3\right)t^3\right]\hat{j} \\ =\left[\left(2 {\text{m/s}}^2\right){\left(10 \text{s}\right)}^2\right]\hat{i}+\left[50 \text{m}+\left(0.5 {\text{m/s}}^3\right){\left(10 \text{s}\right)}^3\right]\hat{j} \\ =\left(200\hat{i}+550\hat{j}\right) \text{m} \end{gathered}$$

Thus, the position vector of the rocket after firing is $\left(200\hat{i}+550\hat{j}\right) \text{m}$