Q44P - University Physics with Modern Physics

Question

A bird flies in the x-y-plane with a velocity vector given by $\overrightarrow{\mathbf{v}}\mathbf{=}\left(\alpha -\beta t^2\right)\hat{\mathbf{i}}\mathbf{+}\mathit{\gamma }\mathit{t}\hat{\mathbf{j}}$  with  $\mathit{\alpha }\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{4}\mathbf{ }\mathit{m}\mathbf{/}\mathit{s}\mathbf{,}\mathit{\beta }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{3}}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }\mathit{\gamma }\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$ , and. The positive y-direction is vertically upward. At t = 0 the bird is at the origin. (a) Calculate the position and acceleration vectors of the bird as functions of time. (b) What is the bird’s altitude (y-coordinate) as it flies over x = 0 for the first time after t = 0?

Step-by-Step Solution

Answer

a) The position and acceleration vector as function of time are $\left(\alpha t+\frac{\beta }{3}{t}^3\right)\hat{i}+\left(\frac{\gamma }{2}{t}^2\right)\hat{j}$ and $\left(-2\beta t\right)\hat{i}+\gamma t\hat{j}$ respectively.

b) the altitude of bird when it flies first time is 

Step 1: Identification of given data

The given data can be listed below,

• The velocity vector of the bird is, $\overrightarrow{v}=\left(\alpha -\beta t^2\right)\hat{i}+\gamma t\hat{j}$
• The value of constants are, $\alpha =2.4m/s,\beta =1.6m/s^3,\gamma =4.0m/s^2$

Step 2: Concept/Significance of altitude.

A reference source and a point or objects are separated by a distance known as altitude, which is often measured vertically or "up” direction.

Step 3: (a) Determination of the position and acceleration vectors of the bird as functions of time

The position vector of bird is calculated by integrating the velocity vector which can be given by,

$\overrightarrow{r}={\int }_0^t\overrightarrow{v}\left(t\right)dt$

Substitute the value in the above,

$$\begin{gathered} \overrightarrow{r}={\int }_0^t\left({\left(\alpha -\beta t\right)}^2\hat{i}+\gamma t\hat{j}\right)dt \\ =\left(\alpha t-\frac{\beta }{3}{t}^3\right)\hat{i}+\left(\frac{\gamma }{2}{t}^2\right)\hat{j} \end{gathered}$$

The acceleration of the bird is calculated by differentiating velocity vector with respect to time which can be given by,

$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}$

Substitute values of velocity vector in the above,

$$\begin{gathered} \overrightarrow{a}=\frac{d\left(\left(\alpha -\beta t^2\right)\hat{i}+\gamma t\hat{j}\right)}{dt} \\ =\left(-2\beta t\right)\hat{i}+\gamma t\hat{j} \end{gathered}$$

Thus, the position and acceleration vector as function of time are $\left(\alpha t-\frac{\beta }{3}{t}^3\right)\hat{i}+\left(\frac{\gamma }{2}{t}^2\right)\hat{j}$ and $\left(-2\beta t\right)\hat{i}+\gamma t\hat{j}$ respectively.

Step4:(b) Determination of the bird’s altitude (y-coordinate) as it flies over x = 0 for the first time after t = 0

The time at which the the horizontal component is zero is given by,

$$\begin{gathered} \alpha t-\frac{\beta }{3}{t}^3=0 \\ t^2=\frac{3\alpha }{\beta } \end{gathered}$$

The altitude of bird is given by,

$y=\frac{\gamma }{2}{t}^2$

Substitute all the values in the above,

$$\begin{gathered} y=\frac{\gamma }{2}\left(\frac{3\alpha }{\beta }\right) \\ =\frac{3\left(2.4 m/s\right)\left(4 m/s^2\right)}{2\left(\left(1.6 m/s^3\right)\right)} \\ =9.0 m \end{gathered}$$

Thus, the altitude of bird when it flies first time is 9.0 m