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Chapter 3: Problem 2

Exercises \(1-6\) give the positions \(s=f(t)\) of a body moving on a coordinate line, with \(s\) in meters and \(t\) in seconds. a. Find the body's displacement and average velocity for the given time interval. b. Find the body's speed and acceleration at the endpoints of the interval. c. When, if ever, during the interval does the body change direction? $$ s=6 t-t^{2}, \quad 0 \leq t \leq 6 $$

Short Answer

Expert verified
1a: Displacement = 0, Average velocity = 0 m/s. 1b: Speed = 6 m/s, Acceleration = -2 m/s虏. 1c: Direction changes at t=3.

Step by step solution

01

Calculate Displacement

To find the body's displacement over the interval \(0 \leq t \leq 6\), we evaluate the position function \(s=f(t)\) at the endpoints of the interval. First, compute \(s(6)\):\[s(6) = 6 \times 6 - 6^2 = 36 - 36 = 0 \]Then, compute \(s(0)\):\[s(0) = 6 \times 0 - 0^2 = 0\]The displacement is the difference in position: \[\text{Displacement} = s(6) - s(0) = 0 - 0 = 0 \]
02

Calculate Average Velocity

The average velocity is defined as the total displacement divided by the total time, which is the interval length. Given the displacement is \(0\) and the interval is \([0, 6]\) seconds, \[\text{Average velocity} = \frac{\text{Displacement}}{\text{Time Interval}} = \frac{0}{6} = 0 \text{ m/s}\]
03

Calculate Speed at the Endpoints

To find the body's speed at the endpoints \(t = 0\) and \(t = 6\), we first find the derivative of \(s(t)\) to get the velocity function:\[v(t) = \frac{ds}{dt} = 6 - 2t\]Evaluate \(v(0)\):\[v(0) = 6 - 2 \times 0 = 6 \text{ m/s}\]The speed is the magnitude of velocity, so the speed at \(t=0\) is \(|6| = 6 \text{ m/s}\).Evaluate \(v(6)\):\[v(6) = 6 - 2 \times 6 = 6 - 12 = -6 \text{ m/s}\]The speed at \(t=6\) is \(|-6| = 6 \text{ m/s}\).
04

Calculate Acceleration at the Endpoints

Acceleration is the derivative of velocity. Differentiate \(v(t) = 6 - 2t\):\[a(t) = \frac{dv}{dt} = -2\]Since the acceleration is constant, it is \(-2 \text{ m/s}^2\) at both endpoints \(t=0\) and \(t=6\).
05

Check for Direction Change

The body changes direction if its velocity changes sign within the interval. This happens when \(v(t) = 0\).Solve \(6 - 2t = 0\):\[2t = 6 \implies t = 3 \]Since \(t=3\) is within the interval \([0, 6]\), the body changes direction at \(t=3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Average Velocity
Average velocity is a fundamental concept in calculus, which describes the overall change in position over a specific time period. It provides a way to assess how fast an object is moving, on average, during a given time interval.
This concept is simple yet powerful in understanding motion.To calculate it, you need to know the total displacement and the time over which this displacement occurs:
  • Displacement is the difference in position at the end and start of the time interval.
  • The time interval is simply the duration over which the motion is considered.
  • The formula for average velocity is: \[ \text{Average velocity} = \frac{\text{Displacement}}{\text{Time Interval}} \]
In our example of the moving body, since the displacement over the interval from 0 to 6 seconds is zero, the average velocity is zero meters per second. This means that, overall, the body didn't gain or lose position across the entire interval.
Acceleration
Acceleration refers to how quickly the velocity of an object is changing. It provides insight into the dynamics of an object's movement, showing if the object is speeding up or slowing down.
Understanding acceleration is key to grasping the bigger picture of motion.The calculation of acceleration involves differentiating the velocity function:
  • Velocity is the rate of change of position with respect to time.
  • Acceleration is thus the derivative of velocity.
  • In mathematical terms: \[ a(t) = \frac{dv}{dt} \]
In the provided problem, the velocity equation is given as \(v(t) = 6 - 2t\). By differentiating \(v(t)\), the acceleration \(a(t)\) is found to be constant at \(-2\, \text{m/s}^2\). This constant acceleration indicates that the velocity decreases steadily over time with no variation.
Direction Change
Direction change occurs when the velocity of an object changes sign. It is a crucial point in the motion of an object as it indicates reversing or "turning back" in motion.
This concept is pivotal for studying motion dynamics and is easy to understand with calculus tools.To determine if direction change happens:
  • The velocity function needs to be set to zero and solved for time.
  • This reveals when the speed is zero and the direction shifts.
  • Formulaically, determine when: \[v(t) = 0\]
  • In our example, the equation \(6 - 2t = 0\) gives \(t = 3\).
Here, the body changes its direction at \(t = 3\) seconds, as within this interval, the velocity becomes zero, switching from positive to negative, indicating that the body reverses its path.

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Most popular questions from this chapter

Use a CAS to perform the following steps. \begin{equation} \begin{array}{l}{\text { a. Plot the equation with the implicit plotter of a CAS. Check to }} \\ {\text { see that the given point } P \text { satisfies the equation. }} \\ {\text { b. Using implicit differentiation, find a formula for the deriva- }} \\ {\text { tive } d y / d x \text { and evaluate it at the given point } P .}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. Use the slope found in part (b) to find an equation for the tan- }} \\ {\text { gent line to the curve at } P \text { . Then plot the implicit curve and }} \\ {\text { tangent line together on a single graph. }}\end{array} \end{equation} \begin{equation} y^{3}+\cos x y=x^{2}, \quad P(1,0) \end{equation}

Use a CAS to perform the following steps. \begin{equation} \begin{array}{l}{\text { a. Plot the equation with the implicit plotter of a CAS. Check to }} \\ {\text { see that the given point } P \text { satisfies the equation. }} \\ {\text { b. Using implicit differentiation, find a formula for the deriva- }} \\ {\text { tive } d y / d x \text { and evaluate it at the given point } P .}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. Use the slope found in part (b) to find an equation for the tan- }} \\ {\text { gent line to the curve at } P \text { . Then plot the implicit curve and }} \\ {\text { tangent line together on a single graph. }}\end{array} \end{equation} \begin{equation} x y^{3}+\tan (x+y)=1, \quad P\left(\frac{\pi}{4}, 0\right) \end{equation}

Quadratic approximations $$ \begin{array}{l}{\text { a. Let } Q(x)=b_{0}+b_{1}(x-a)+b_{2}(x-a)^{2} \text { be a quadratic }} \\ {\text { approximation to } f(x) \text { at } x=a \text { with the properties: }}\end{array} $$ $$ \begin{aligned} \text { i) } Q(a) &=f(a) \\ \text { ii) } Q^{\prime}(a) &=f^{\prime}(a) \\ \text { iii) } Q^{\prime \prime}(a) &=f^{\prime \prime}(a) \end{aligned} $$ $$ \text{Determine the coefficients}b_{0}, b_{1}, \text { and } b_{2} $$ $$ \begin{array}{l}{\text { b. Find the quadratic approximation to } f(x)=1 /(1-x) \text { at }} \\ {\quad x=0 .} \\ {\text { c. } \operatorname{Graph} f(x)=1 /(1-x) \text { and its quadratic approximation at }} \\ {x=0 . \text { Then zoom in on the two graphs at the point }(0,1) \text { . }} \\ {\text { Comment on what you see. }}\end{array} $$ $$ \begin{array}{l}{\text { d. Find the quadratic approximation to } g(x)=1 / x \text { at } x=1} \\ {\text { Graph } g \text { and its quadratic approximation together. Comment }} \\ {\text { on what you see. }}\end{array} $$ $$ \begin{array}{l}{\text { e. Find the quadratic approximation to } h(x)=\sqrt{1+x} \text { at }} \\ {x=0 . \text { Graph } h \text { and its quadratic approximation together. }} \\ {\text { Comment on what you see. }} \\\ {\text { f. What are the linearizations of } f, g, \text { and } h \text { at the respective }} \\ {\text { points in parts (b), (d), and (e)? }}\end{array} $$

If \(x^{2}+y^{2}=25\) and \(d x / d t=-2,\) then what is \(d y / d t\) when \(x=3\) and \(y=-4 ?\)

Use a CAS to perform the following steps. \begin{equation} \begin{array}{l}{\text { a. Plot the equation with the implicit plotter of a CAS. Check to }} \\ {\text { see that the given point } P \text { satisfies the equation. }} \\ {\text { b. Using implicit differentiation, find a formula for the deriva- }} \\ {\text { tive } d y / d x \text { and evaluate it at the given point } P .}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. Use the slope found in part (b) to find an equation for the tan- }} \\ {\text { gent line to the curve at } P \text { . Then plot the implicit curve and }} \\ {\text { tangent line together on a single graph. }}\end{array} \end{equation} \begin{equation} x+\tan \left(\frac{y}{x}\right)=2, \quad P\left(1, \frac{\pi}{4}\right) \end{equation}
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