Problem 44 - University Physics with Modern Physics

Question

A bird flies in the \(xy\)-plane with a velocity vector given by \(\overrightarrow{v}\) = \((a - \beta$$t^{2})\) \(\hat{l}+\gamma t \hat{j}\), with \(\alpha = 2.4 m/s\), \(\beta = 1.6 m/s^3\), and \(\gamma = 4.0 m/s^2\). 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

Position: \(\overrightarrow{r} = (\alpha t - \frac{\beta t^3}{3}) \hat{i} + \frac{\gamma t^2}{2} \hat{j}\) Acceleration: \(\overrightarrow{a} = (-3.2 t) \hat{i} + 4.0 \hat{j}\) Altitude at \(x=0\) (after \(t=0\)): 18 m.

Step 1: Understanding the Given Problem

We need to find the position and acceleration vectors of the bird given its velocity as \(\overrightarrow{v} = (\alpha - \beta t^2) \hat{i} + \gamma t \hat{j}\), and identify when it passes \(x=0\). Then, determine the altitude when it crosses \(x=0\) for the first time after \(t=0\).

Step 2: Calculate the Acceleration Vector

The acceleration vector \(\overrightarrow{a}\) is the derivative of the velocity vector \(\overrightarrow{v}\) with respect to time \(t\). Differentiating with respect to \(t\), we have:\[\overrightarrow{a} = \dfrac{d}{dt}((\alpha - \beta t^2) \hat{i} + \gamma t \hat{j})\]This gives us:\[\overrightarrow{a} = (-2\beta t) \hat{i} + \gamma \hat{j}\]Substituting \(\beta = 1.6\, \text{m/s}^3\) and \(\gamma = 4.0\, \text{m/s}^2\), we get:\[\overrightarrow{a} = (-3.2 t) \hat{i} + 4.0 \hat{j}\].

Step 3: Calculate the Position Vector

To find the position vector \(\overrightarrow{r}\), integrate the velocity vector \(\overrightarrow{v}\) with respect to time \(t\).The integration of \((\alpha - \beta t^2)\) with respect to \(t\) gives:\[x(t) = \int (\alpha - \beta t^2) dt = \alpha t - \dfrac{\beta t^3}{3} + C_1\]Integration of \(\gamma t\) gives:\[y(t) = \int \gamma t dt = \dfrac{\gamma t^2}{2} + C_2\]Since the bird starts at the origin \((0,0)\) when \(t=0\), \(C_1=0\) and \(C_2=0\). Therefore, we have:\[x(t) = \alpha t - \dfrac{\beta t^3}{3}\]\[y(t) = \dfrac{\gamma t^2}{2}\].

Step 4: Determine the First Instance Bird is Directly Above x=0

We need the time when \(x(t) = \alpha t - \dfrac{\beta t^3}{3} = 0\) again, apart from \(t=0\).Solving:\[\alpha t = \dfrac{\beta t^3}{3}\]\[t(\alpha - \dfrac{\beta t^2}{3}) = 0\]Excluding the solution \(t=0\), we solve:\[\alpha - \dfrac{\beta t^2}{3} = 0\]\[\dfrac{\beta t^2}{3} = \alpha\]\[t^2 = \dfrac{3\alpha}{\beta}\]Calculating gives \(t = \sqrt{\dfrac{3 \cdot 2.4}{1.6}}\), which simplifies to \(t = 3\) seconds.

Step 5: Calculate Altitude Above x=0 at t=3

Using \(t=3\) seconds in the equation for \(y(t)\):\[y(3) = \dfrac{\gamma \cdot 3^2}{2} = \dfrac{4 \cdot 9}{2} = 18 \text{ m}\].Therefore, the bird's altitude when it passes over \(x=0\) for the first time after \(t=0\) is 18 meters.

Key Concepts

Velocity Vector

In kinematics, a velocity vector describes the speed and direction of an object's movement. For a bird flying in the xy-plane, as given in our problem, the velocity vector is expressed as \( \overrightarrow{v} = (\alpha - \beta t^2) \hat{i} + \gamma t \hat{j} \).
This shows that the velocity of the bird consists of two components:

• The \( \hat{i} \) component: \( (\alpha - \beta t^2) \) indicating velocity in the horizontal \( x \)-direction.
• The \( \hat{j} \) component: \( \gamma t \) indicating velocity in the vertical \( y \)-direction.

Understanding each part of this vector helps decipher how the bird's speed changes over time in both horizontal and vertical directions.
The horizontal component decreases with increasing time due to \( -\beta t^2 \), while the vertical component increases linearly over time due to \( \gamma t \). This suggests the bird's path is being affected by increasing energy in the vertical upward movement and a decreasing tendency in the horizontal direction as it moves.

Acceleration Vector

The acceleration vector, \( \overrightarrow{a} \), represents how the velocity of the bird changes over time. It is derived by taking the derivative of the velocity vector with respect to time \( t \).
Given the velocity vector \( \overrightarrow{v} = (\alpha - \beta t^2) \hat{i} + \gamma t \hat{j} \),
its time derivative will be:

• \( \overrightarrow{a} = \frac{d}{dt}((\alpha - \beta t^2) \hat{i} + \gamma t \hat{j}) \)
• Resulting in \( \overrightarrow{a} = (-2\beta t) \hat{i} + \gamma \hat{j} \).

The acceleration vector thus reveals:

• A negative component \( -2\beta t \) in the \( \hat{i} \) direction, which implies a slowing down in the horizontal direction over time.
• A constant upward component \( \gamma \) in the \( \hat{j} \) direction, indicating a steady increase in vertical speed.

Plugging in the values \( \beta = 1.6 \) and \( \gamma = 4 \), we get \( \overrightarrow{a} = (-3.2t) \hat{i} + 4 \hat{j} \), which numerically confirms these trends.

Position Vector

The position vector \( \overrightarrow{r} \) provides the actual position of the bird at any given time. It is found by integrating the velocity vector \( \overrightarrow{v} \) over time.

The velocities in the x and y directions are integrated separately:

• For the x-direction, integrate \( \alpha - \beta t^2 \) with respect to time:
  \[ x(t) = \int (\alpha - \beta t^2)\,dt = \alpha t - \frac{\beta t^3}{3} + C_1 \]
• For the y-direction, integrate \( \gamma t \):
  \[ y(t) = \int \gamma t\,dt = \frac{\gamma t^2}{2} + C_2 \]

Since the bird starts at the origin \((0,0)\) at \( t=0 \), constants \( C_1 \) and \( C_2 \) are both zero.
Hence, the position functions are:
\[ x(t) = \alpha t - \frac{\beta t^3}{3} \]
\[ y(t) = \frac{\gamma t^2}{2} \]
This means that as time evolves, the bird moves horizontally according to a polynomial effect, while its vertical motion is quadratically increasing.

Integrating Velocity

Integrating velocity allows us to compute an object's position from its velocity-time function. When the velocity vector \( \overrightarrow{v} = (\alpha - \beta t^2) \hat{i} + \gamma t \hat{j} \) is integrated, we obtain the position vector \( \overrightarrow{r} \).

Integration translates the rate of change of a quantity (velocity) into the actual quantity (position).

• The integration of \( \alpha - \beta t^2 \) over time results in \( x(t) = \alpha t - \frac{\beta t^3}{3} \), showcasing a polynomial decrease due to \( -\frac{\beta t^3}{3} \).
• The term \( \gamma t \), when integrated, results in \( y(t) = \frac{\gamma t^2}{2} \), showing how vertical distance evolves through squaring time.

Using conditions like initial positions allows simplifying constants \( C_1 \) and \( C_2 \) to zero.
In essence, integrating velocity defines how displacement accumulates over time, providing the complete trajectory of the bird from its velocities.