Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

If a particle moves along a coordinate line so that its directed distance from the origin after \(t\) seconds is \(\left(-t^{2}+4 t\right)\) feet, when did the particle come to a momentary stop (i.e., when did its instantaneous velocity become zero)?

Step-by-Step Solution

Answer

The particle stops momentarily at \(t = 2\) seconds.

Step 1: Identify the Position Function

The position of the particle is given by the function \(s(t) = -t^2 + 4t\). This represents the distance the particle is from the origin at any given time \(t\).

Step 2: Find the Velocity Function

The instantaneous velocity is found by taking the derivative of the position function with respect to time \(t\). So, the velocity function is \(v(t) = \frac{d}{dt}(-t^2 + 4t)\).

Step 3: Differentiate the Position Function

Differentiating \(-t^2 + 4t\) gives \(v(t) = -2t + 4\). This is the velocity function of the particle.

Step 4: Set the Velocity Function to Zero

To find when the particle comes to a momentary stop, set the velocity function \(v(t) = 0\): \(-2t + 4 = 0\).

Step 5: Solve for Time \(t\)

Solve the equation \(-2t + 4 = 0\) for \(t\) to find when the velocity is zero.\[ -2t + 4 = 0 \Rightarrow -2t = -4 \Rightarrow t = 2 \]

Step 6: Interpret the Solution

The particle comes to a momentary stop at \(t = 2\) seconds when the instantaneous velocity is zero.

Key Concepts

Understanding the Position Function

When a particle moves along a straight line, its location can be described using the position function. In the context of the given exercise, the position function is denoted as \(s(t) = -t^2 + 4t\). This equation tells us where the particle is relative to the origin at any time \(t\). The expression \(-t^2 + 4t\) is derived from applying the rules of algebra to represent how far the particle has traveled over time.

• The term \(-t^2\) suggests that the movement includes a squared time component, indicating acceleration or deceleration.
• Meanwhile, the term \(4t\) represents a linear change over time.

By substituting different values of \(t\) in the equation, we can determine the exact position of the particle at those times.

Calculating the Velocity Function

The velocity function helps us understand how fast the particle's position is changing over time. To find this function, we need to differentiate the position function with respect to time \(t\). This is because velocity is the rate at which position changes. Taking the derivative:

• The differentiation of \(s(t) = -t^2 + 4t\) gives us \(v(t) = \frac{d}{dt}(-t^2 + 4t)\).
• This results in the velocity function \(v(t) = -2t + 4\).

This velocity function \(v(t) = -2t + 4\) shows us how the speed and direction of the particle vary as time progresses. Higher complexity in the position function, such as higher degrees of \(t\) or non-linear terms, would similarly influence the complexity of the velocity expression.

Delving into Differentiation

Differentiation is an essential tool in calculus used to compute the rate of change of a function. In this exercise, we use differentiation to find the velocity function from the position function.

• When you differentiate a function like \(-t^2 + 4t\), you apply the power rule, which simplifies each term of the polynomial.
• For instance, the derivative of \( -t^2 \) is \(-2t\) because the power rule decreases the exponent by one and multiplies the term by the original exponent.
• For the linear component \(4t\), the derivative is simply \(4\).

Understanding the differentiation process makes it easier to transition from knowing where an object is (its position) to understanding how fast it is moving (its velocity).

Examining Instantaneous Velocity

Instantaneous velocity refers to the speed of an object at a specific moment in time. It is a concept focused on what happens over an "infinitely small" time period.In the exercise, we find the instantaneous velocity to be zero when the function \(-2t + 4\) equals zero. This is because at these moments, the particle is neither moving forwards nor backwards, indicating a stop.

• By setting \(v(t) = 0\), we get the equation \(-2t + 4 = 0\), and solving this for \(t\) gives us \(t = 2\).
• At \(t = 2\) seconds, the instantaneous velocity hits zero, signifying a momentary stop.

The concept of instantaneous velocity helps in understanding how objects can stop and start or change direction, revealing insights into motion that might not be obvious from just looking at average speeds.