Problem 16 - Calculus Early Transcendentals

Question

The displacement (in meters) of a particle moving in a straight line is given by \(s=t^{2}-8 t+18\) , where \(t\) is measured in seconds. (a) Find the average velocity over each time interval: $$\begin{array}{ll}{\text { (i) }[3,4]} & {\text { (ii) }[3.5,4]} \\ {\text { (iii) }[4,5]} & {\text { (iv) }[4,4.5]}\end{array}$$ (b) Find the instantaneous velocity when \(t=4\) (c) Draw the graph of s as a function of t and draw the secant lines whose slopes are the average velocities in part (a) and the tangent line whose slope is the instantaneous velocity in nart (b)

Step-by-Step Solution

Answer

(a) Velocities: -1, -0.5, 1, 0.5 m/s (b) Instantaneous velocity is 0 m/s.

Step 1: Displacement Function Evaluation

To find the average velocity over a time interval, we first evaluate the displacement function at the endpoints of that interval. The displacement function given is \( s = t^2 - 8t + 18 \). Calculate \( s(t) \) for each endpoint in the intervals:- For \( t = 3 \), \( s(3) = 3^2 - 8 \times 3 + 18 = 9 - 24 + 18 = 3 \).- For \( t = 4 \), \( s(4) = 4^2 - 8 \times 4 + 18 = 16 - 32 + 18 = 2 \).- For \( t = 3.5 \), \( s(3.5) = (3.5)^2 - 8 \times 3.5 + 18 = 12.25 - 28 + 18 = 2.25 \).- For \( t = 5 \), \( s(5) = 5^2 - 8 \times 5 + 18 = 25 - 40 + 18 = 3 \).- For \( t = 4.5 \), \( s(4.5) = (4.5)^2 - 8 \times 4.5 + 18 = 20.25 - 36 + 18 = 2.25 \).

Step 2: Calculate Average Velocities

The formula for average velocity \( v_{avg} \) over a time interval \([t_1, t_2]\) is \( v_{avg} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \). Calculate for each interval:- For \([3,4]\), \( v_{avg} = \frac{2 - 3}{4 - 3} = \frac{-1}{1} = -1 \text{ m/s} \).- For \([3.5,4]\), \( v_{avg} = \frac{2 - 2.25}{4 - 3.5} = \frac{-0.25}{0.5} = -0.5 \text{ m/s} \).- For \([4,5]\), \( v_{avg} = \frac{3 - 2}{5 - 4} = \frac{1}{1} = 1 \text{ m/s} \).- For \([4,4.5]\), \( v_{avg} = \frac{2.25 - 2}{4.5 - 4} = \frac{0.25}{0.5} = 0.5 \text{ m/s} \).

Step 3: Find Instantaneous Velocity

The instantaneous velocity is the derivative of the displacement function evaluated at a specific time. First, find the derivative of \( s(t) = t^2 - 8t + 18 \):\[ \frac{ds}{dt} = 2t - 8 \]Evaluate this at \( t = 4 \):\[ v_{inst} = 2(4) - 8 = 8 - 8 = 0 \text{ m/s} \].

Step 4: Graph the Displacement and Velocities

Plot the graph of \( s(t) = t^2 - 8t + 18 \) with time \( t \) on the x-axis and displacement \( s(t) \) on the y-axis.- Draw secant lines for each interval: 1. From (3,3) to (4,2) for \([3,4]\). 2. From (3.5,2.25) to (4,2) for \([3.5,4]\). 3. From (4,2) to (5,3) for \([4,5]\). 4. From (4,2) to (4.5,2.25) for \([4,4.5]\).- Draw a tangent line at \( t = 4 \) with slope \( 0 \), representing the instantaneous velocity.

Key Concepts

Displacement Function

A displacement function describes how the position of an object changes over time. In our given exercise, the displacement function is represented by the equation \( s = t^2 - 8t + 18 \). This equation lets us calculate the position, or displacement, of the particle at any given time \( t \). By substituting different values of \( t \) into the function, we determine where the particle is at those specific times. This information is crucial for accurately determining velocity, as it allows us to track how far the particle has moved over intervals of time.

Instantaneous Velocity

Instantaneous velocity is a measure of how fast an object is moving at a specific moment in time. It can be thought of as the speed and direction at that precise instant. To find the instantaneous velocity, we calculate the derivative of the displacement function with respect to time. The derivative gives us a new function that represents the rate of change of position with time.
In the case of our displacement function \( s = t^2 - 8t + 18 \), the derivative is \( \frac{ds}{dt} = 2t - 8 \). By substituting \( t = 4 \) into this derivative, we find that the instantaneous velocity at \( t = 4 \) seconds is \( 0 \text{ m/s} \). This result implies that at that moment, the particle is neither moving forward nor backward, suggesting a pause or a momentary change of direction.

Secant and Tangent Lines

Secant and tangent lines are geometric tools used to understand average and instantaneous rates of change in functions. A secant line intersects a curve at two points and its slope represents the average rate of change over an interval. In our exercise, each computed average velocity corresponds to the slope of a different secant line drawn between two points on the displacement graph.

• For the interval \([3,4]\), the secant line's slope is \(-1 \text{ m/s} \).
• For \([3.5,4]\), it is \(-0.5 \text{ m/s} \).
• For \([4,5]\), it is \(1 \text{ m/s} \).
• Finally, for \([4,4.5]\), the slope is \(0.5 \text{ m/s} \).

The tangent line, however, touches the curve at only one point and its slope represents the instantaneous rate of change. At \( t = 4 \), the tangent line has a slope of \( 0 \), indicating the particle's instantaneous velocity at that moment is \( 0 \text{ m/s} \). This contrast between secant and tangent lines vividly illustrates the difference between average and instantaneous velocity.