Problem 119 - Fundamentals of Physics

Question

The position of a particle as it moves along a \(y\) axis is given by $$ y=(2.0 \mathrm{~cm}) \sin (\pi t / 4) $$ with \(t\) in seconds and \(y\) in centimeters. (a) What is the average velocity of the particle between \(t=0\) and \(t=2.0 \mathrm{~s} ?\) (b) What is the instantaneous velocity of the particle at \(t=0,1.0,\) and \(2.0 \mathrm{~s} ?\) (c) What is the average acceleration of the particle between \(t=0\) and \(t=2.0 \mathrm{~s} ?\) (d) What is the instantancous acceleration of the particle at \(t=0\), \(1.0,\) and \(2.0 \mathrm{~s} ?\)

Step-by-Step Solution

Answer

Average velocity: 1.0 cm/s. Instantaneous velocities: \(\frac{\pi}{2}\) cm/s (at 0s), \(\frac{\pi\sqrt{2}}{4}\) cm/s (at 1.0s), 0 cm/s (at 2.0s). Average acceleration: \(-\frac{\pi}{4}\) cm/s². Instantaneous accelerations: 0 cm/s² (at 0s), \(-\frac{\pi^2\sqrt{2}}{16}\) cm/s² (at 1.0s), \(-\frac{\pi^2}{8}\) cm/s² (at 2.0s).

Step 1: Given Function

The position of a particle is given by the equation \( y = (2.0 \text{ cm}) \sin(\pi t / 4) \), where \( t \) is in seconds and \( y \) is in centimeters.

Step 2: Average Velocity Formula

The formula for average velocity \( v_{avg} \) over a time interval \([t_1, t_2]\) is \( v_{avg} = \frac{y(t_2) - y(t_1)}{t_2 - t_1} \).

Step 3: Calculate Positions at Key Times for Average Velocity

First, find \( y(t) \) at \( t=0 \) and \( t=2.0 \).\[ y(0) = (2.0 \text{ cm}) \sin(\pi \cdot 0 / 4) = 0 \text{ cm} \] \[ y(2.0) = (2.0 \text{ cm}) \sin(\pi \cdot 2 / 4) = 2.0 \text{ cm} \]

Step 4: Compute Average Velocity

With the positions calculated, the average velocity between \( t=0 \) and \( t=2.0 \) seconds is: \[ v_{avg} = \frac{y(2.0) - y(0)}{2.0 - 0} = \frac{2.0 \text{ cm} - 0 \text{ cm}}{2.0 \text{ s}} = 1.0 \text{ cm/s} \]

Step 5: Instantaneous Velocity Derivative

To find the instantaneous velocity \( v(t) \), differentiate \( y(t) = (2.0 \text{ cm}) \sin(\pi t / 4) \) with respect to \( t \): \[ v(t) = \frac{dy}{dt} = (2.0 \text{ cm}) \cdot \frac{\pi}{4} \cos(\pi t / 4) = \frac{\pi}{2} \cos(\pi t / 4) \text{ cm/s} \]

Step 6: Calculate Instantaneous Velocities

Substitute the values of \( t \) into the velocity equation: - At \( t=0 \): \[ v(0) = \frac{\pi}{2} \cos(0) = \frac{\pi}{2} \text{ cm/s} \] - At \( t=1.0 \): \[ v(1.0) = \frac{\pi}{2} \cos(\pi / 4) = \frac{\pi}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\pi \sqrt{2}}{4} \text{ cm/s} \] - At \( t=2.0 \): \[ v(2.0) = \frac{\pi}{2} \cos(\pi / 2) = 0 \text{ cm/s} \]

Step 7: Average Acceleration Formula

The formula for average acceleration \( a_{avg} \) over a time interval \([t_1, t_2]\) is \( a_{avg} = \frac{v(t_2) - v(t_1)}{t_2 - t_1} \).

Step 8: Compute Average Acceleration

\[ a_{avg} = \frac{v(2.0) - v(0)}{2.0 - 0} = \frac{0 \text{ cm/s} - \frac{\pi}{2} \text{ cm/s}}{2.0 \text{ s}} = -\frac{\pi}{4} \text{ cm/s}^2 \]

Step 9: Instantaneous Acceleration Derivative

To find the instantaneous acceleration \( a(t) \), differentiate the velocity function: \[ a(t) = \frac{d^2y}{dt^2} = - \frac{\pi^2}{8} \sin(\pi t / 4) \text{ cm/s}^2 \]

Step 10: Calculate Instantaneous Accelerations

Substitute the values of \( t \) into the acceleration equation: - At \( t=0 \): \[ a(0) = - \frac{\pi^2}{8} \sin(0) = 0 \text{ cm/s}^2 \] - At \( t=1.0 \): \[ a(1.0) = - \frac{\pi^2}{8} \sin(\pi / 4) = - \frac{\pi^2 \sqrt{2}}{16} \text{ cm/s}^2 \] - At \( t=2.0 \): \[ a(2.0) = - \frac{\pi^2}{8} \sin(\pi / 2) = - \frac{\pi^2}{8} \text{ cm/s}^2 \]

Key Concepts

Average Velocity

Average velocity describes how fast a particle is moving overall during a specific time frame. In essence, it tells you the total change in position divided by the total time taken. For the given exercise, we focus on the time interval from \(t=0\) to \(t=2.0\) seconds, using the formula:

• \( v_{avg} = \frac{y(t_2) - y(t_1)}{t_2 - t_1} \)

Here, \(y(t)\) is the position function. First, we need to find the particle's position at these times. Given that \(y(t) = (2.0 \text{ cm}) \sin(\pi t / 4)\), at \(t=0\), \(y(0) = 0 \text{ cm}\) and at \(t=2.0\), \(y(2.0) = 2.0 \text{ cm}\). Therefore:

• Average velocity \(v_{avg} = \frac{2.0 \text{ cm} - 0 \text{ cm}}{2.0 \text{ s}} = 1.0 \text{ cm/s}\)

This means the particle has moved at an average speed of \(1.0 \text{ cm/s}\) in this timeframe.

Instantaneous Velocity

Instantaneous velocity indicates how fast and in which direction a particle is traveling at a specific moment in time. It's the rate of change of position at a particular instant. To find this, we need to take the derivative of the position function \(y(t)\) with respect to time \(t\):

• \( v(t) = \frac{dy}{dt} = \frac{\pi}{2} \cos(\pi t / 4) \text{ cm/s} \)

Now, let's evaluate it at certain moments:

• At \(t=0\): \( v(0) = \frac{\pi}{2} \cdot \cos(0) = \frac{\pi}{2} \text{ cm/s}\)
• At \(t=1.0\): \( v(1.0) = \frac{\pi}{2} \cdot \cos(\pi / 4) = \frac{\pi \sqrt{2}}{4} \text{ cm/s}\)
• At \(t=2.0\): \( v(2.0) = \frac{\pi}{2} \cdot \cos(\pi / 2) = 0 \text{ cm/s}\)

This shows that at \(t=0\) the particle moves at a speed of \(\frac{\pi}{2} \text{ cm/s}\), slows down at \(t=1.0\), and comes to a stop at \(t=2.0\).

Average Acceleration

Average acceleration refers to the overall change in velocity over a period of time. It gives us an idea of how much the speed or direction of the particle is changing on average. The formula for average acceleration is:

• \( a_{avg} = \frac{v(t_2) - v(t_1)}{t_2 - t_1} \)

From our previous calculations, we know the velocity at \(t=0\) is \(\frac{\pi}{2} \text{ cm/s}\) and at \(t=2.0\) is \(0 \text{ cm/s}\). Substituting these values, we find:

• Average acceleration \(a_{avg} = \frac{0 \text{ cm/s} - \frac{\pi}{2} \text{ cm/s}}{2.0 \text{ s}} = -\frac{\pi}{4} \text{ cm/s}^2\)

This negative sign indicates the particle is decelerating, meaning its velocity is decreasing over the time interval from \(t=0\) to \(t=2.0\).

Instantaneous Acceleration

Instantaneous acceleration depicts how quickly a particle's velocity is changing at any given moment. To find this, differentiate the instantaneous velocity function with respect to time:

• \( a(t) = \frac{d^2y}{dt^2} = - \frac{\pi^2}{8} \sin(\pi t / 4) \text{ cm/s}^2 \)

Now, let's calculate it at specific times:

• At \(t=0\): \( a(0) = - \frac{\pi^2}{8} \cdot \sin(0) = 0 \text{ cm/s}^2\)
• At \(t=1.0\): \( a(1.0) = - \frac{\pi^2}{8} \cdot \sin(\pi / 4) = - \frac{\pi^2 \sqrt{2}}{16} \text{ cm/s}^2\)
• At \(t=2.0\): \( a(2.0) = - \frac{\pi^2}{8} \cdot \sin(\pi / 2) = - \frac{\pi^2}{8} \text{ cm/s}^2\)

These values explain how the particle's velocity is changing instantaneously. At \(t=0\), the acceleration is zero, meaning no change in velocity at that instant. As \(t\) progresses, the particle experiences acceleration, affecting the speed and direction of its motion.