**$\overrightarrow r =(\alpha t – \beta t^3)\hat{i}+\gamma t^2\hat{j}$****$\alpha =2.4\dfrac{m}{s}$****$\beta=1.6\dfrac{m}{s^3}$****$\gamma=4.0\dfrac{m}{s^2}$****Compute the bird’s acceleration vector as a function of time.****What is the altitude y-coordinate of the bird when it first flies to x = 0?**

This **task** aims to find the velocity and acceleration **vectors of** a bird moving **within xy-plane using the** position vector **specified** in the question. The average acceleration vector is defined as the rate of change in velocity, or the** direction** in **which** the **speed changes. ****Velocity**, on the other hand, is the rate of **change of displacement**. The velocity vector v always points in the **direction of motion**.

## Expert Answer

**(a)** The **direction** of the $y-axis$ is **vertically upward**. Bird is at the origin at $t=0$. The **velocity vector **$(v=\dfrac{dr}{dt})$ is obtained by the **derivative of position vector** with **respect to time**.

\[\overrightarrow v =(\alpha t – 3\beta t^2)\overrightarrow i+2\gamma t^1\overrightarrow j\]

\[\overrightarrow v =(2.4t – 4.8t^2)\overrightarrow i+8.0t\overrightarrow j\]

**(b)** The **acceleration vector** is the **derivative** of **velocity vector** with respect to **time**.

\[a(t)=\dfrac{dv(t)}{dt}\]

\[\overrightarrow a =(-6\beta t)\overrightarrow i+2\gamma \overrightarrow j\]

\[\overrightarrow a=(-9.6t)\overrightarrow i+8.0\overrightarrow j\]

**(c) **First, find the time when the $x$ component of the** position vector** is equal to **zero**.

\[\alpha t- \dfrac{\beta t^3}{3}=0\]

\[\alpha=\dfrac{\beta t^3}{3}\]

\[t=\sqrt {\dfrac{3\alpha}{\beta}}=2.12s\]

**Plug** these values into the $y-component$.

\[y(t)=\dfrac{\gamma t^2}{2}\]

\[y(2.12)=\dfrac{4(2.12)^2}{2}=9m\]

## Numerical Results

**(a)** Velocity vector of the bird as the function of time is:

\[\overrightarrow v =(2.4t – 4.8t^2)\overrightarrow i+8.0t\overrightarrow j\]

**(b)** **Acceleration vector** of the **bird as the function of time** is:

\[\overrightarrow a=(-9.6t)\overrightarrow i+8.0\overrightarrow j\]

**(c) Bird altitude** when $x$-component is **zero**.

\[y(2.12)=\dfrac{4(2.12)^2}{2}=9m\]

## Example

A bird flies in the $xy$-plane with a position vector given by $\overrightarrow r =(\alpha t – \beta t^3)\hat{i}+\gamma t^2\hat{j}$ , with $\alpha =4.4\dfrac{m}{s}$, $\beta=2\dfrac{m}{s^3}$, and $\gamma=6.0\dfrac{m}{s^2}$.The positive $y$-direction is vertically upward. At the bird is at the origin.

**-Calculate the velocity vector of the bird as a function of time.**

**-Compute the bird’s acceleration vector as a function of time.**

**-What is the altitude $(y\:coordinate)$ of the bird when it first flies to $x = 0$?**

**(a)** The **direction** of the $y-axis$ is **vertically upward**. Bird is at the origin at $t=0$. The **velocity vector** is function of time $(v=\dfrac{dr}{dt})$.The **velocity vector** is obtained by **derivative of position vector** with **respect to time**.

\[\overrightarrow v =(\alpha t – 3\beta t^2)\overrightarrow i+2\gamma t^1\overrightarrow j\]

**Velocity vector** is given as:

\[\overrightarrow v =(4.4t – 6t^2)\overrightarrow i+12.0t\overrightarrow j\]

**(b)** The **acceleration vector** is the **derivative** of **velocity vector** with respect to **time**.

\[a(t)=\dfrac{dv(t)}{dt}\]

\[\overrightarrow a =(-6\beta t)\overrightarrow i+2\gamma \overrightarrow j\]

Thus, **acceleration vector** is given as:

\[\overrightarrow a=(-12t)\overrightarrow i+12.0\overrightarrow j\]

**(c) **First, find the time when the $x$ component of the** position vector** is equal to **zero**.

\[\alpha t- \dfrac{\beta t^3}{3}=0\]

\[\alpha=\dfrac{\beta t^3}{3}\]

\[t=\sqrt {\dfrac{3\alpha}{\beta}}=2.6s\]

**Plug** these values into the $y-component$.

\[y(t)=\dfrac{\gamma t^2}{2}\]

\[y(2.12)=\dfrac{6(2.6)^2}{2}=20.2m\]

Thus, **altitude** is $20.2m$ across $y$-axis