Problem 4 - The Calculus with Analytic Geometry

Question

A particle is moving along a horizontal line according to the given equation of motion, where \(s \mathrm{ft}\) is the directed distance of the particle from a point \(O\) at \(t \mathrm{sec}\). Find the instantaneous velocity \(v\left(t_{1}\right) \mathrm{ft} / \mathrm{sec}\) at \(t_{1} \mathrm{sec}\); and then find \(v\left(t_{1}\right)\) for the particular value of \(t_{1}\) given. $$ s=8-t^{2} ; t_{1}=5 $$

Step-by-Step Solution

Answer

The instantaneous velocity at \( t = 5 \) is \( -10 \text{ ft/sec} \).

Step 1: Understand the Problem

The particle's position as a function of time is given by the equation of motion: \[ s = 8 - t^2 \]We need to find the instantaneous velocity at a particular time, specifically when \( t = 5 \).

Step 2: Differentiate the Position Function

Velocity is the derivative of position with respect to time. Differentiate \( s \) with respect to \( t \):\[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(8 - t^2) \]Apply the power rule for differentiation:\[ v(t) = 0 - 2t = -2t \]

Step 3: Substitute the Given Time Value

Substitute \( t_1 = 5 \) into the velocity function to find the instantaneous velocity:\[ v(t_1) = -2(5) = -10 \]Therefore, the instantaneous velocity at \( t = 5 \) seconds is \[ -10 \text{ ft/sec} \].

Key Concepts

Equation of Motion

In physics and mathematics, an equation of motion describes the position of a particle or object with respect to time. It often takes the form of a function, such as \( s(t) \), representing the position \( s \) at any given time \( t \). Understanding the equation of motion is crucial because it helps us determine where an object is at any point in time. In our exercise, the position function is given as:

\[ s = 8 - t^2 \]This tells us that the particle's distance from a reference point changes with time according to the square of the time subtracted from a constant value. Here, \( s \) is measured in feet, and \( t \) in seconds.

• When \( t = 0 \), the particle starts 8 feet away from the reference point.
• As \( t \) increases, the distance \( s \) decreases because of the \( -t^2 \) term.

Usually, analyzing an equation of motion involves various steps, including differentiation to find velocities or accelerations.

Derivative

A derivative represents the rate at which one quantity changes with respect to another. In the context of an equation of motion, the derivative of the position function gives us the velocity. The velocity tells us how fast the position of the particle is changing at any given moment.
Let's differentiate the given position function \( s = 8 - t^2 \):

• The derivative \( \frac{ds}{dt} \) is computed by differentiating each term in the position function.
• For \( 8 \), which is a constant, the derivative is 0 because constants don’t change.
• For \( -t^2 \), we differentiate using the power rule (to be discussed next) to get \( -2t \).

Putting it all together, the velocity function \( v(t) \) is: \[ v(t) = \frac{ds}{dt} = -2t \]

Power Rule for Differentiation

The power rule is one of the most fundamental rules in differentiation. It states that for any function \( x^n \), where \( n \) is a constant, the derivative is given by bringing down the exponent as a coefficient and reducing the exponent by one:

\[ \frac{d}{dx} x^n = n x^{n-1} \]This rule simplifies the process of differentiation significantly. Applying this to our position function:

• Identify that \( t^2 \) is of the form \( x^n \) where \( n=2 \).
• Use the power rule: bring down the 2 as a coefficient and reduce the exponent by one.

This gives us: \[ \frac{d}{dt} (-t^2) = -2t \]Therefore, differentiating \( 8 - t^2 \) results in the velocity function \( v(t) = -2t \).

Finally, to find the instantaneous velocity at \( t = 5 \), we substitute \( t = 5 \) into the velocity expression:

\[ v(t_1) = -2 \times 5 = -10 \text{ ft/sec} \]
This negative sign indicates the particle is moving in the opposite direction relative to the positive coordinate direction.