Problem 15 - College Physics

Question

A mouse travels along a straight line; its distance \(x\) from the origin at any time \(t\) is given by the equation \(x=\) \(\left(8.5 \mathrm{~cm} \cdot \mathrm{s}^{-1}\right) t-\left(2.5 \mathrm{~cm} \cdot \mathrm{s}^{-2}\right) t^{2} .\) Find the average velocity of the mouse in the interval from \(t=0\) to \(t=1.0 \mathrm{~s}\) and in the interval from \(t=0\) to \(t=4.0 \mathrm{~s}\)

Step-by-Step Solution

Answer

The average velocity is 6.0 cm/s for the first interval and -1.5 cm/s for the second interval.

Step 1: Understand the Formula for Average Velocity

The average velocity is defined as the change in position (displacement) divided by the time interval during which the change occurs. Mathematically, it is given by: \[ \text{Average Velocity} = \frac{x(t_2) - x(t_1)}{t_2 - t_1} \] where \( x(t_2) \) and \( x(t_1) \) are the positions at times \( t_2 \) and \( t_1 \), respectively.

Step 2: Calculate Positions at Given Times

First, we substitute the given times into the equation to find the mouse's position.For \( t = 0 \): \[ x(0) = (8.5 \, \text{cm} \, \text{s}^{-1})\cdot 0 - (2.5 \, \text{cm} \, \text{s}^{-2})\cdot 0^2 = 0 \, \text{cm} \]For \( t = 1.0 \, \text{s} \): \[ x(1) = (8.5 \, \text{cm} \, \text{s}^{-1})\cdot 1 - (2.5 \, \text{cm} \, \text{s}^{-2})\cdot 1^2 = 8.5 \, \text{cm} - 2.5 \, \text{cm} = 6.0 \, \text{cm} \]For \( t = 4.0 \, \text{s} \): \[ x(4) = (8.5 \, \text{cm} \, \text{s}^{-1})\cdot 4 - (2.5 \, \text{cm} \, \text{s}^{-2})\cdot 4^2 = 34 \, \text{cm} - 40 \, \text{cm} = -6 \, \text{cm} \]

Step 3: Calculate Average Velocity for Each Interval

Now, we calculate the average velocity for the specified intervals using the positions we found:For \( t=0 \) to \( t=1.0 \, \text{s} \):\[ \text{Average Velocity} = \frac{x(1) - x(0)}{1.0 - 0} = \frac{6.0 \, \text{cm} - 0 \, \text{cm}}{1.0 \, \text{s}} = 6.0 \, \text{cm/s} \]For \( t=0 \) to \( t=4.0 \, \text{s} \):\[ \text{Average Velocity} = \frac{x(4) - x(0)}{4.0 - 0} = \frac{-6.0 \, \text{cm} - 0 \, \text{cm}}{4.0 \, \text{s}} = -1.5 \, \text{cm/s} \]

Key Concepts

Displacement

Displacement refers to the change in position of an object from its starting point to its endpoint. Unlike distance, displacement considers the direction of movement and is defined along a straight line connecting the start and end points.
In this exercise, we calculate the displacement of a mouse traveling along a straight line. If the mouse’s position at time \( t_1 \) is \( x(t_1) \) and its position at time \( t_2 \) is \( x(t_2) \), then the displacement \( \Delta x \) is given by the equation: - \[ \Delta x = x(t_2) - x(t_1) \] This indicates how far and in which direction the mouse has traveled from its initial position.
When the displacement is positive, the mouse moves forward; when negative, it moves backward. Displacement is a vector quantity, which means it has both magnitude and direction.

Time Interval

The time interval is the total amount of time over which an event occurs. It is the difference between the final time \( t_2 \) and the initial time \( t_1 \).
To calculate it, use the formula: - \[ \Delta t = t_2 - t_1 \] In our scenario, for instance, the time interval from \( t=0 \) to \( t=1.0 \) seconds is simply \( 1.0 - 0 = 1.0 \, \text{s} \). Similarly, for the interval from \( t=0 \) to \( t=4.0 \) seconds, \( \Delta t = 4.0 - 0 = 4.0 \, \text{s} \).
The time interval is crucial when calculating the average velocity, as it determines how long the mouse traveled for each segment of its journey. Understanding the concept of time interval helps in planning and analyzing motion more effectively.

Position

Position describes the location of an object at a particular time. It's determined by using an equation of motion that represents how the position changes over time.
In our example, the position \( x(t) \) of the mouse is defined by the equation: - \( x = (8.5 \,\text{cm} \cdot \text{s}^{-1})t - (2.5 \,\text{cm} \cdot \text{s}^{-2})t^2 \) Here, \((8.5 \,\text{cm} \cdot \text{s}^{-1})\) is the initial velocity, and \((2.5 \,\text{cm} \cdot \text{s}^{-2})\) is the acceleration of the mouse. This equation allows us to find the position of the mouse at any point in time by substituting \( t \) with the required time value.
A key step in solving average velocity problems is to determine the position at the start and end of each interval to find the necessary displacement.

Velocity Calculation

The calculation of velocity involves finding the speed of an object with direction. Specifically, average velocity is measured by considering the total displacement of an object over the total time interval of travel.
Mathematically, average velocity \( v_a \) is expressed through the formula: - \[ v_a = \frac{\Delta x}{\Delta t} \] For example, if during the interval from \( t=0 \) to \( t=1.0 \) seconds, the mouse's displacement \( \Delta x \) is \( 6.0 \, \text{cm} \), and \( \Delta t \) is \( 1.0 \, \text{s} \), the average velocity would be \( \frac{6.0 \, \text{cm}}{1.0 \, \text{s}} = 6.0 \, \text{cm/s} \).
This indicates that the mouse moved in one second at a rate of 6.0 cm/s during that time frame. Similarly, other intervals can be calculated to understand the motion dynamics.