Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

An object travels along a line so that its position \(s\) is \(s=t^{2}+1\) meters after \(t\) seconds. (a) What is its average velocity on the interval \(2 \leq t \leq 3\) ? C (b) What is its average velocity on the interval \(2 \leq t \leq 2.003\) ? (c) What is its average velocity on the interval \(2 \leq t \leq 2+h\) ? (d) Find its instantaneous velocity at \(t=2\).

Step-by-Step Solution

Answer

Average velocities: (a) 5 m/s, (b) 4.003 m/s, (c) 4 + h m/s. Instantaneous velocity at \(t=2\): 4 m/s.

Step 1: Understanding Average Velocity

The average velocity of an object over a time interval is defined as the total displacement divided by the total time taken. It is given by the formula:\[\text{Average velocity} = \frac{s(t_{2}) - s(t_{1})}{t_{2} - t_{1}}\]where \(s(t)\) is the position function.

Step 2: Calculating Displacement for Part (a)

First, find the positions at \(t=2\) and \(t=3\). The position when \(t=2\) is:\[s(2) = 2^2 + 1 = 5\]The position when \(t=3\) is:\[s(3) = 3^2 + 1 = 10\]Displacement is \(s(3) - s(2) = 10 - 5 = 5\) meters.

Step 3: Calculating Average Velocity for Part (a)

Using the displacement and time interval \(3 - 2 = 1\) second, we compute the average velocity:\[\text{Average velocity} = \frac{5}{1} = 5\text{ m/s}\]

Step 4: Calculating Displacement for Part (b)

Find the position at \(t=2.003\). The position is:\[s(2.003) = (2.003)^2 + 1 = 4.012009 + 1 = 5.012009\]Displacement for \([2, 2.003]\) is \(5.012009 - 5 = 0.012009\) meters.

Step 5: Calculating Average Velocity for Part (b)

The time interval for part (b) is \(2.003 - 2 = 0.003\) seconds. The average velocity is:\[\text{Average velocity} = \frac{0.012009}{0.003} = 4.003\text{ m/s}\]

Step 6: Understanding the General Case for Part (c)

For the interval \([2, 2+h]\), use the position difference and time difference:\[\text{Average velocity} = \frac{s(2+h) - s(2)}{h}\]

Step 7: Calculating Average Velocity for Part (c)

Find the expression for position at \(2+h\):\[s(2+h) = (2+h)^2 + 1 = (4 + 4h + h^2) + 1 = 5 + 4h + h^2\]Thus:\[\text{Average velocity} = \frac{(5 + 4h + h^2) - 5}{h} = 4 + h\]

Step 8: Finding Instantaneous Velocity for Part (d)

To find the instantaneous velocity, take the derivative of the position function \(s(t) = t^2 + 1\):\[\frac{ds}{dt} = 2t\]Evaluate at \(t=2\):\[\frac{ds}{dt}\Big|_{t=2} = 2(2) = 4\text{ m/s}\]

Key Concepts

Average Velocity

Average velocity is an essential concept in calculus that describes how fast an object is moving along a straight line over a given time interval. To calculate it, we need to find the total displacement and divide it by the elapsed time. In mathematical terms, this is expressed as: \[\text{Average velocity} = \frac{s(t_{2}) - s(t_{1})}{t_{2} - t_{1}}.\]This formula uses a position function, where \(s(t)\) represents the location of the object at time \(t\).

For example, if an object’s position is described by the equation \(s=t^2+1\), the average velocity on the interval \(2 \leq t \leq 3\) can be calculated by finding the displacement, which is \(s(3) - s(2) = 10 - 5 = 5\) meters, and dividing it by the time difference \((3-2)\), resulting in an average velocity of \(5\ \text{m/s}\).

This same concept applies to smaller intervals like \(2 \leq t \leq 2.003\), where the displacement \(0.012009\) over a time \(0.003\) results in an average velocity of approximately \(4.003\ \text{m/s}\).

Instantaneous Velocity

Instantaneous velocity, unlike average velocity, focuses on the speed of an object at a single point in time rather than over an interval. It is the velocity at which an object is moving at exactly one moment, effectively the limit of the average velocity as the time interval becomes infinitesimally small.

To determine instantaneous velocity, we use calculus and specifically derivatives. By differentiating the position function \(s(t) = t^2 + 1\), we can find the rate of change at any point \(t\).

For the position function given, the derivative is \(\frac{ds}{dt} = 2t\). At \(t = 2\), the instantaneous velocity is \(2 \times 2 = 4\ \text{m/s}\), indicating that at 2 seconds, the object moves at 4 meters per second.

Derivative

Derivatives play a crucial role in understanding motion in calculus. They provide the rate at which a function is changing, translating geometrically to the slope of the tangent line to the curve representing the function.

When considering motion, the derivative of the position function with respect to time gives the instantaneous velocity, capturing both speed and direction at any given point in time.

In the position function \(s(t) = t^2 + 1\), the derivative is \(\frac{ds}{dt} = 2t\). This tells us how the object’s position changes with every passing second. By evaluating this derivative at a specific time, such as \(t = 2\), we get the instantaneous velocity, showing that derivatives not only predict future motion but also describe immediate behavior.

Position Function

A position function expresses where an object is located along a line at any given time. In the context of motion, it models the journey over time and is fundamental for predicting future positions, finding velocities, and understanding motion patterns.

In our example, the position function is \(s(t) = t^2 + 1\), which means that for any value of time \(t\), the position \(s(t)\) can be easily calculated. For instance, at \(t = 2\), \(s(2) = 5\), and at \(t = 3\), \(s(3) = 10\).

These position readings are crucial for computing displacements, determining both average and instantaneous velocities, and illustrating the progression of motion. Understanding a position function allows us to grasp not only where an object is or will be but also how it got there.