Problem 4 - University Physics with Modern Physics

Question

CALC The position of a squirrel running in a park is given by \(\vec{r}= [(0.280 m/s)t + (0.0360 m/s^2)t^2] \hat{\imath}+(0.0190 m/s^3)t^3\hat{\jmath}\). (a) What are \(v_x(t)\) and \(v_y(t)\), the \(x\)- and \(y\)-components of the velocity of the squirrel, as functions of time? (b) At \(t\) = 5.00 s, how far is the squirrel from its initial position? (c) At \(t\) = 5.00 s, what are the magnitude and direction of the squirrel's velocity?

Step-by-Step Solution

Answer

(a) \(v_x(t) = 0.280 + 0.0720t\), \(v_y(t) = 0.0570t^2\). (b) Distance: 3.31 m. (c) Velocity: 1.56 m/s at 65.67° from x-axis.

Step 1: Find the x-component of the velocity

To find the x-component of the velocity \( v_x(t) \), differentiate the x-component of the position vector \( (0.280 \, m/s)t + (0.0360 \, m/s^2)t^2 \) with respect to time \( t \).\[ v_x(t) = \frac{d}{dt}[(0.280 \, m/s)t + (0.0360 \, m/s^2)t^2] = 0.280 \, m/s + 2(0.0360 \, m/s^2)t = 0.280 \, m/s + 0.0720 \, m/s^2 \cdot t \]

Step 2: Find the y-component of the velocity

To find the y-component of the velocity \( v_y(t) \), differentiate the y-component of the position vector \( (0.0190 \, m/s^3)t^3 \) with respect to time \( t \). \[ v_y(t) = \frac{d}{dt}[(0.0190 \, m/s^3)t^3] = 3(0.0190 \, m/s^3)t^2 = 0.0570 \, m/s^3 \cdot t^2 \]

Step 3: Calculate the position at t = 5.00 s

Substitute \( t = 5.00 \, s \) into the position vector equation to calculate the squirrel's position. For the x-component:\[ x(5.00) = (0.280 \, m/s)(5.00) + (0.0360 \, m/s^2)(5.00)^2 = 1.40 \, m + 0.90 \, m = 2.30 \, m \] For the y-component:\[ y(5.00) = (0.0190 \, m/s^3)(5.00)^3 = 2.375 \, m \]

Step 4: Calculate displacement from initial position

The initial position is \( \vec{r}(0) = 0 \). The displacement is the absolute value of the change in position vector.Displacement is given by:\[ \sqrt{(2.30 \, m)^2 + (2.375 \, m)^2} = \sqrt{5.29 + 5.640625} = \sqrt{10.930625} \approx 3.306 \text{ m} \]

Step 5: Find the velocity components at t = 5.00 s

Using equations from Steps 1 and 2, calculate \( v_x(5.00) \) and \( v_y(5.00) \).\[ v_x(5.00) = 0.280 \, m/s + 0.0720 \, m/s^2 \times 5.00 = 0.640 \, m/s \]\[ v_y(5.00) = 0.0570 \, m/s^3 \times (5.00)^2 = 1.425 \, m/s \]

Step 6: Calculate magnitude and direction of velocity at t = 5.00 s

The magnitude of the velocity is given by:\[ v = \sqrt{(v_x)^2 + (v_y)^2} = \sqrt{(0.640 \, m/s)^2 + (1.425 \, m/s)^2} = \sqrt{0.4096 + 2.030625} = \sqrt{2.440225} \approx 1.562 \text{ m/s} \]To find the direction, use:\[ \theta = \tan^{-1} \left( \frac{v_y}{v_x} \right) = \tan^{-1} \left( \frac{1.425}{0.640} \right) \approx 65.67^\circ \]

Key Concepts

Velocity Components

When a squirrel runs through the park, its movement can be described by breaking its velocity into components along the horizontal, or x-axis, and the vertical, or y-axis. These are called velocity components.
In our exercise, the squirrel's position over time is given by two equations: one for the x-direction and one for the y-direction. These equations can be differentiated to find the velocity components:

• x-component: The equation for the position in the x-direction is \( (0.280 \, m/s)t + (0.0360 \, m/s^2)t^2 \). By differentiating this equation with respect to time \( t\), we find the velocity in the x-direction: \( v_x(t) = 0.280 \, m/s + 0.0720 \, m/s^2 \cdot t \).
• y-component: The position in the y-direction is \( (0.0190 \, m/s^3)t^3 \). Differentiating this gives us the velocity in the y-direction: \( v_y(t) = 0.0570 \, m/s^3 \cdot t^2 \).

Illustrating this notion helps us understand how the squirrel’s speed changes in each direction as it moves.

Differentiation in Physics

Differentiation, a fundamental concept in calculus, is also immensely important in physics. It's the process used to find rates of change, such as velocity from position.
In the position equation of the squirrel, differentiation allows us to derive the velocity of the squirrel. For example, the expression for the x-component of the squirrel's position, \( (0.280 \, m/s)t + (0.0360 \, m/s^2)t^2 \), is differentiated with respect to time \( t\) to give the x-component of velocity, \( v_x(t)\).

• Differentiating \( t \) terms results in each term being one degree lower in order. The coefficient also modifies according to the power rule of differentiation, which is \( d(ax^n)/dx = nax^{n-1} \).

This concept provides the bridge from static analysis of positions to dynamic descriptions involving speed.

Magnitude and Direction

Understanding an object's overall velocity requires analyzing both the magnitude and the direction.
The magnitude is the speed at which the squirrel is moving, regardless of direction. It is calculated using the Pythagorean theorem, combining the velocity components. If \( v_x \) and \( v_y \) are known, the velocity magnitude \( v \) is found by: \[ v = \sqrt{(v_x)^2 + (v_y)^2} \]
The direction specifies the line along which the object is moving. It's determined using trigonometry, specifically the tangent function. For our squirrel, the direction \( \theta \) is:

• Calculated by \( \theta = \tan^{-1} \left( \frac{v_y}{v_x} \right) \)

This angle tells us how much the squirrel is veering from the horizontal axis in its path. By grasping these concepts, one can describe motion both in terms of speed and path orientation.