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A bird flies in the xy-plane with a position vector given below .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.

  • $\overrightarrow r =(\alpha t – \beta t^3)\hat{i}+\gamma t^2\hat{j}$
  • $\alpha =2.4\dfrac{m}{s}$
  • $\beta=1.6\dfrac{m}{s^3}$
  • $\gamma=4.0\dfrac{m}{s^2}$
  • 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

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