Problem 15 - Calculus Early Transcendentals

Question

The displacement (in meters) of a particle moving in a straight line is given by the equation of motion \(s=1 / t^{2}\) where \(t\) is measured in seconds. Find the velocity of the particle at times \(t=a, t=1, t=2,\) and \(t=3\) .

Step-by-Step Solution

Answer

Velocity at \( t=a, t=1, t=2, t=3 \) is \(-\frac{2}{a^3}, -2, -\frac{1}{4}, -\frac{2}{27}\) respectively.

Step 1: Define the Relationship Between Displacement and Velocity

In general, velocity is the rate of change of displacement with respect to time. To find the velocity, we need to differentiate the displacement function, which is given by: \[ s = \frac{1}{t^2} \].

Step 2: Differentiate the Displacement Function

To find the velocity function \( v(t) \), we differentiate \( s(t) = \frac{1}{t^2} \) with respect to \( t \). Using the power rule for differentiation, where \( \frac{d}{dt} \left(t^{-n}\right) = -nt^{-n-1} \), we get: \[ v(t) = \frac{d}{dt} \left( t^{-2} \right) = -2t^{-3} = -\frac{2}{t^3} \].

Step 3: Calculate Velocity at Specific Times

Substitute the given times into the velocity function to find the velocity at each point. - For \( t = a \): \[ v(a) = -\frac{2}{a^3} \].- For \( t = 1 \): \[ v(1) = -\frac{2}{1^3} = -2 \].- For \( t = 2 \): \[ v(2) = -\frac{2}{2^3} = -\frac{2}{8} = -\frac{1}{4} \].- For \( t = 3 \): \[ v(3) = -\frac{2}{3^3} = -\frac{2}{27} \].

Key Concepts

Displacement Function

In physics, understanding the concept of displacement is crucial when analyzing the movement of objects. Displacement refers to the overall change in position of an object from its starting point. It's a vector quantity, meaning it has both a magnitude and a direction. In the given problem, the displacement of a particle moving in a straight line is expressed by the function:

• \( s(t) = \frac{1}{t^2} \)

This function describes how the position of the particle changes over time, where \( s \) is the displacement in meters, and \( t \) is the time in seconds.
The form \( \frac{1}{t^2} \) signifies that as time increases, the displacement decreases due to the particle moving closer to the starting point. This is a common form for inverse relationships, where one quantity increases as another decreases. To fully grasp the displacement function, it's beneficial to visualize it, perhaps by plotting a graph which might exhibit a curve that approaches zero as time moves forward.

Velocity Function

Velocity is a fundamental concept when discussing motion. It describes how quickly position changes with time. In mathematical terms, velocity is the derivative of the displacement function with respect to time. Differentiation is a core concept in calculus that allows us to find the rate at which a quantity changes.
For the displacement function \( s(t) = \frac{1}{t^2} \), we differentiate with respect to \( t \) to find the velocity function, \( v(t) \). By applying the power rule for differentiation, we obtain:

• \( v(t) = -\frac{2}{t^3} \)

This new function gives us the velocity of the particle at any point in time. It's essential to recognize that this velocity function includes a negative sign, indicating that the particle's speed is decreasing over time, as it slows down while moving in the reverse direction relative to time.

Rate of Change

The rate of change is an essential concept in understanding how one quantity varies relative to another. In the context of motion, it's often expressed as the derivative of a function. When dealing with displacement and velocity, the rate of change gives insight into how the position of an object changes over time.

• The derivative of the displacement function \( s(t) = \frac{1}{t^2} \) results in the velocity function \( v(t) = -\frac{2}{t^3} \).

This process, differentiation, shows that the rate at which the displacement decreases is determined by \( v(t) \). More concretely, for specific values of \( t \), such as \( t = 1, 2, \) or \( 3 \), we can calculate the velocity to understand the instantaneous rate of displacement change at those moments.
For example:

• At \( t = 1 \), the velocity is \( -2 \), meaning the particle moves 2 meters per second in the negative direction.
• At \( t = 2 \), the velocity is \(-\frac{1}{4}\), a much slower rate of change.

This illustrates how differentiation informs the understanding of motion dynamics, providing a clear view of how speed and direction evolve over time.