CC Determining Distance Traveled from Velocity

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Section 4.1 Determining Distance Traveled from Velocity

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Motivating Questions

If we know the velocity of a moving body at every point in a given interval, can we determine the distance the object has traveled on that time interval?
How is the problem of finding distance traveled related to finding the area under a certain curve?
What does it mean to antidifferentiate a function and why is this process relevant to finding distance traveled?
If velocity is negative, how does this impact the problem of finding distance traveled?

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Supplemental Videos.

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The main topics of this section are also presented in the following videos:

Estimating Distance Traveled
¹
unl.yuja.com/V/Video?v=7114339&node=34303218&a=8858494&autoplay=1
Refining Estimates to Determine Exact Distance
²
unl.yuja.com/V/Video?v=7114338&node=34303299&a=65860470&autoplay=1
Area Beneath Velocity Curve Represents Distance
³
unl.yuja.com/V/Video?v=7114335&node=34303245&a=711360&autoplay=1

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In the first chapter of the text, we considered a moving object with known position at time

t,

namely, a tennis ball tossed into the air with height

s

(in feet) at time

t

(in seconds) given by

s (t) = 64 - 16 (t - 1)^{2} .

We investigated the average velocity of the ball on an interval

[a, b],

computed by the difference quotient

\frac{s (b) - s (a)}{b - a} .

We found that we could determine the instantaneous velocity of the ball at time

t

by taking the derivative of the position function,

s^{'} (t) = lim_{h \to 0} \frac{s (t + h) - s (t)}{h} .

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Thus, if its position function is differentiable, we can find the velocity of a moving object at any point in time.

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From this study of position and velocity we have learned a great deal. We can use the derivative to find a function’s instantaneous rate of change at any point in the domain, to find where the function is increasing or decreasing, where it is concave up or concave down, and to locate relative extremes. The vast majority of the problems and applications we have considered have involved the situation where a particular function is known and we seek information that relies on knowing the function’s instantaneous rate of change. For all these tasks, we proceed from a function

f

to its derivative,

f^{'},

and use the meaning of the derivative to help us answer important questions.

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We have also encountered the reverse situation, where we know the derivative of a function,

f^{'},

and try to deduce information about

f .

We will focus our attention in Chapter 4 on this problem: if we know the instantaneous rate of change of a function, can we find the function itself? We start with a more specific question: if we know the instantaneous velocity of an object moving along a straight line path, can we find its corresponding position function?

Example 4.1.

Suppose that a person is taking a walk along a long straight path and walks at a constant rate of 3 miles per hour.

Sketch and label a graph of the velocity function $v (t) = 3 .$
How far did the person travel during the two hours? How is this distance related to the area of a certain region under the graph of $y = v (t) ?$
Find an algebraic formula, $s (t),$ for the position of the person at time $t,$ assuming that $s (0) = 0 .$ Explain your thinking.
Sketch a labeled graph of the position function $y = s (t) .$
For what values of $t$ is the position function $s$ increasing? Explain why this is the case using relevant information about the velocity function $v .$

Hint.

Let $t$ be the input and $v$ be the output.
Think about how you would calculate how far you travel in a certain amount of time while traveling at a constant speed.
Use the same method as in part (b), but for a general value of $t .$
Let $t$ be the input and $s$ be the output.
Use the graph you drew.

Answer.

Figure 4.2. The velocity is constant, which creates a horizontal line.
The person traveled 6 miles in 2 hours. This is the same as the area under the curve of $v (t)$ between 0 and 2.
$s (t) = 3 t$
Figure 4.3. The position function.
The position function $s (t)$ is always increasing. This is because the velocity is constantly positive.

Solution.

Figure 4.4. The velocity is constant, which creates a horizontal line.
If the person is traveling at a constant speed of 3 miles per hour, we can find the distance traveled by multiplying the speed by the amount of time they are walking. So, the person traveled 6 miles in 2 hours.

The height of the function is always at 3 and the time is given by the $x$ -axis. The distance traveled is the same as the area under the curve of $v (t)$ between 0 and 2.
For every time, the position is given by multiplying the constant velocity, 3, by the time. Therefore, $s (t) = 3 t .$
Figure 4.5. The position function.
The position function $s (t)$ is always increasing. This is because the velocity is constantly positive. We can see this also by looking at the graph of $s (t),$ which is a straight line with a positive slope.

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Subsection 4.1.1 Area under the graph of the velocity function

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In Example 4.1, we learned that when the velocity of a moving object’s velocity is constant (and positive), the area under the velocity curve over an interval of time tells us the distance the object traveled.

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Figure 4.6. At left, a constant velocity function; at right, a non-constant velocity function.

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The left-hand graph of Figure 4.6 shows the velocity of an object moving at 2 miles per hour over the time interval

[1, 1.5] .

The area

A_{1}

of the shaded region under

y = v (t)

[1, 1.5]

A_{1} = 2 \frac{miles}{hour} \cdot \frac{1}{2} hours = 1 mile .

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This result is simply the fact that distance equals rate times time, provided the rate is constant. Thus, if

v (t)

is constant on the interval

[a, b],

the distance traveled on

[a, b]

is equal to the area

A

given by

A = v (a) (b - a) = v (a) Δ t,

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where

Δ t

is the change in

t

over the interval. (Since the velocity is constant, we can use any value of

v (t)

on the interval

[a, b],

we simply chose

v (a),

the value at the interval’s left endpoint.) For several examples where the velocity function is piecewise constant, see http://gvsu.edu/s/9T

⁴

gvsu.edu/s/9T

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Marc Renault, calculus applets.

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The situation is more complicated when the velocity function is not constant. But on relatively small intervals where

v (t)

does not vary much, we can use the area principle to estimate the distance traveled. The right-hand graph of Figure 4.6 shows a non-constant velocity function. On the interval

[1, 1.5],

the velocity varies from

v (1) = 2.5

down to

v (1.5) \approx 2.1 .

One estimate for the distance traveled is the area of the pictured rectangle,

A_{2} = v (1) Δ t = 2.5 \frac{miles}{hour} \cdot \frac{1}{2} hours = 1.25 miles .

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Note that because

v

is decreasing on

[1, 1.5],

A_{2} = 1.25

is an over-estimate of the actual distance traveled.

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To estimate the area under this non-constant velocity function on a wider interval, say

[0, 3],

one rectangle will not give a good approximation. Instead, we could use the six rectangles pictured in Figure 4.7, find the area of each rectangle, and add up the total. Obviously there are choices to make and issues to understand: How many rectangles should we use? Where should we evaluate the function to decide the rectangle’s height? What happens if the velocity is sometimes negative? Can we find the exact area under any non-constant curve?

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Figure 4.7. Using six rectangles to estimate the area under $y = v (t)$ on $[0, 3] .$

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We will study these questions and more in what follows; for now it suffices to observe that the simple idea of the area of a rectangle gives us a powerful tool for estimating distance traveled from a velocity function, as well as for estimating the area under an arbitrary curve. To explore the use of multiple rectangles to approximate area under a non-constant velocity function, see the applet found at http://gvsu.edu/s/9U

⁶

gvsu.edu/s/9U

⁷

Marc Renault, calculus applets.

Example 4.8.

Suppose that a person is walking in such a way that her velocity varies slightly according to the information given in Table 4.9 and graph given in Figure 4.10.

$t$	$v (t)$
$0.00$	$1.500$
$0.25$	$1.789$
$0.50$	$1.938$
$0.75$	$1.992$
$1.00$	$2.000$
$1.25$	$2.008$
$1.50$	$2.063$
$1.75$	$2.211$
$2.00$	$2.500$

Table 4.9. Velocity data for the person walking.

Using the grid, graph, and given data, estimate the distance traveled by the walker during the two hour interval from $t = 0$ to $t = 2 .$ You should use time intervals of width $Δ t = 0.5 .$ Choose a way to use the function consistently to determine the height of each rectangle in order to approximate distance traveled.
How could you get a better approximation of the distance traveled on $[0, 2] ?$ Explain, and then find this new estimate.
Suppose that you know that $v$ is given by $v (t) = 0.5 t^{3} - 1.5 t^{2} + 1.5 t + 1.5 .$ Remember that $v$ is the derivative of the walker’s position function, $s .$ Find a formula for $s$ so that $s^{'} = v .$
Based on your work in (c), what is the value of $s (2) - s (0) ?$ What is the meaning of this quantity?

Hint.

For instance, the approximate distance traveled on $[0, 0.5]$ can be computed by $v (0) \cdot 0.5 = 1.5 \cdot 0.5 = 0.75$ miles.
Think about the possibility of using a larger number of rectangles.
If $v (t) = t^{3}$ and we seek a function $s$ such that $s^{'} = v,$ then $s$ has to involve $t^{4} .$
Observe that this quantity is measuring a change in position.

Answer.

$\begin{aligned} A = & v (0.0) \cdot 0.5 + v (0.5) \cdot 0.5 + v (1.0) \cdot 0.5 + v (1.5) \cdot 0.5 \\ = & 1.500 \cdot 0.5 + 1.9375 \cdot 0.5 + 2.000 \cdot 0.5 + 2.0625 \cdot 0.5 \\ = & 3.75 \end{aligned}$

Thus, $D \approx 3.75$ miles.
Using 8 rectangles of width $0.25,$ $D \approx 3.875 .$
$s (t) = \frac{1}{8} t^{4} - \frac{1}{2} t^{3} + \frac{3}{4} t^{2} + \frac{3}{2} t .$
$s (2) - s (0) = (\frac{1}{8} (2)^{4} - \frac{1}{2} (2)^{3} + \frac{3}{4} (2)^{2} + \frac{3}{2} (2)) - (0) = 4 .$

Solution.

Using rectangles of width $Δ t = 0.5$ and choosing to set the heights of the rectangles from the function value at the left end of the interval, we see the following graph and find the sum of the areas of the rectangles to be

$\begin{aligned} A = & v (0.0) \cdot 0.5 + v (0.5) \cdot 0.5 + v (1.0) \cdot 0.5 + v (1.5) \cdot 0.5 \\ = & 1.500 \cdot 0.5 + 1.9375 \cdot 0.5 + 2.000 \cdot 0.5 + 2.0625 \cdot 0.5 \\ = & 3.75 \end{aligned}$

Thus, the distance traveled is approximately $D \approx 3.75$ miles.
It appears that a better approximation could be found using narrower rectangles. If we move to 8 rectangles of width $0.25,$ similar computations show that $D \approx 3.875 .$
By thinking about how the power rule for differentiation works, we can undo this rule and find a position function $s$ whose derivative is $v .$ For instance, since $\frac{d}{d t} [t^{4}] = 4 t^{3},$ we see that $\frac{d}{d t} [\frac{1}{8} t^{4}] = \frac{1}{2} t^{3} .$ Thus, if we let

$s (t) = \frac{1}{8} t^{4} - \frac{1}{2} t^{3} + \frac{3}{4} t^{2} + \frac{3}{2} t,$

then it is straightforward to check that $s^{'} (t) = \frac{1}{2} t^{3} - \frac{3}{2} t^{2} + \frac{3}{2} t + \frac{3}{2},$ which is precisely the formula for $v (t)$ that we were given.
By the rule found in (c) for $s,$ we have that $s (2) - s (0) = \frac{1}{8} 2^{4} - \frac{1}{2} 2^{3} + \frac{3}{4} 2^{2} + \frac{3}{2} 2 = 2 - 4 + 3 + 3 = 4 .$ This is the change in the walker’s position over the time interval $[0, 2],$ and since the velocity is always positive, this is actually the exact distance traveled. We see how both earlier estimates ( $3.75$ and $3.785$ ) are good approximations to this value.

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Subsection 4.1.2 Two approaches: area and antidifferentiation

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When the velocity of a moving object is positive, the object’s position is always increasing. (We will soon consider situations where velocity is negative; for now, we focus on the situation where velocity is always positive.) We have established that whenever

v

is constant on an interval, the exact distance traveled is the area under the velocity curve. When

v

is not constant, we can estimate the total distance traveled by finding the areas of rectangles that approximate the area under the velocity curve.

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Thus, we see that finding the area between a curve and the horizontal axis is an important exercise: besides being an interesting geometric question, if the curve gives the velocity of a moving object, the area under the curve tells us the exact distance traveled on an interval. We can estimate this area if we have a graph or a table of values for the velocity function.

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In Example 4.8, we encountered an alternate approach to finding the distance traveled. If

y = v (t)

is a formula for the instantaneous velocity of a moving object, then

v

must be the derivative of the object’s position function,

s .

If we can find a formula for

s (t)

from the formula for

v (t),

we will know the position of the object at time

t,

and the change in position over a particular time interval tells us the distance traveled on that interval.

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For a simple example, consider the situation from Example 4.1, where a person is walking along a straight line with velocity function

v (t) = 3

mph.

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Figure 4.11. The velocity function $v (t) = 3$ and corresponding position function $s (t) = 3 t .$

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On the left-hand graph of the velocity function in Figure 4.11, we see the relationship between area and distance traveled,

A = 3 \frac{miles}{hour} \cdot 1.25 hours = 3.75 miles .

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In addition, we observe

⁸

Here we are making the implicit assumption that

s (0) = 0;

we will discuss different possibilities for values of

s (0)

in subsequent study.

that if

s (t) = 3 t,

then

s^{'} (t) = 3,

s (t) = 3 t

is the position function whose derivative is the given velocity function,

v (t) = 3 .

The respective locations of the person at times

t = 0.25

and

t = 1.5

are

s (1.5) = 4.5

and

s (0.25) = 0.75,

and therefore

s (1.5) - s (0.25) = 4.5 - 0.75 = 3.75 miles .

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This is the person’s change in position on

[0.25, 1.5],

which is precisely the distance traveled. In this example there are profound ideas and connections that we will study throughout Chapter 4.

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For now, observe that if we know a formula for a velocity function

v,

it can be very helpful to find a function

s

that satisfies

s^{'} = v .

We say that

s

is an antiderivative of

v .

More generally, we have the following formal definition.

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Definition 4.12.

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g

and

G

are functions such that

G^{'} = g,

we say that

G

is an antiderivative of

g .

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For example, if

g (x) = 3 x^{2} + 2 x,

G (x) = x^{3} + x^{2}

is an antiderivative of

g,

because

G^{'} (x) = g (x) .

Note that we say “an” antiderivative of

g

rather than “the” antiderivative of

g,

because

H (x) = x^{3} + x^{2} + 5

is also a function whose derivative is

g,

and thus

H

is another antiderivative of

g .

Example 4.13.

A ball is tossed vertically in such a way that its velocity function is given by

v (t) = 32 - 32 t,

where

t

is measured in seconds and

v

in feet per second. Assume that this function is valid for

0 \leq t \leq 2 .

For what values of $t$ is the velocity of the ball positive? What does this tell you about the motion of the ball on this interval of time values?
Find an antiderivative, $s,$ of $v$ that satisfies $s (0) = 0 .$
Compute the value of $s (1) - s (\frac{1}{2}) .$ What is the meaning of the value you find?
Using the graph of $y = v (t)$ provided in Figure 4.14, find the exact area of the region under the velocity curve between $t = \frac{1}{2}$ and $t = 1 .$ What is the meaning of the value you find?

Figure 4.14. The graph of $y = v (t) .$
Answer the same questions as in (c) and (d) but instead using the interval $[0, 1] .$
What is the value of $s (2) - s (0) ?$ What does this result tell you about the flight of the ball? How is this value connected to the provided graph of $y = v (t) ?$ Explain.

Hint.

Where is velocity zero?
Since $v$ is linear, note that $s$ must be quadratic.
Observe that you are taking the difference between two values of the position function.
The region whose area is sought is triangular.
See (c) and (d) above.
What does it mean for the change of the ball’s position to be zero?

Answer.

On $(0, 1),$ $s$ is increasing because velocity is positive.
$s (t) = 32 t - 16 t^{2} .$
$s (1) - s (\frac{1}{2}) = 4 .$
$A = 4$ feet is the total distance the ball traveled vertically on $[0, \frac{1}{2}] .$
$s (1) - s (0) = 16$ is the vertical distance the ball traveled on the interval $[0, 1] .$ Equivalently, the area under the velocity curve on $[0, 1]$ is $A = 16$ feet.
$s (2) - s (0) = 0,$ so the ball has zero change in position on the interval $[0, 2] .$

Solution.

Note that $v (1) = 0$ and for $0 < t < 1,$ $v (t) > 0 .$ This means that on the interval $(0, 1),$ the position function $s$ is increasing because velocity is positive.
We can check that the derivative of $s (t) = 32 t - 16 t^{2}$ is $s^{'} (t) = v (t) = 32 - 32 t,$ and that $s (0) = 0,$ so this is the antiderivative of $v$ that we desire.
Now, $s (1) - s (\frac{1}{2}) = (32 - 16) - (16 - 4) = 4,$ which is the change in position of the ball on the interval $[\frac{1}{2}, 1] .$ Equivalently, since $v$ is positive through this interval, 4 feet is the vertical distance the ball traveled during this time.
On the interval from $t = \frac{1}{2}$ to $t = 1,$ the corresponding area under the velocity curve is the area of the right triangular region whose width is $\frac{1}{2}$ seconds and whose height is $v (\frac{1}{2}) = 16$ feet/sec. That area is therefore $A = \frac{1}{2} b h = \frac{1}{2} \cdot \frac{1}{2} \cdot 16 = 4$ feet. This is the total distance the ball traveled vertically on $[0, \frac{1}{2}] .$
$s (1) - s (0) = (32 - 16) - (0 - 0) = 16,$ which is the vertical distance the ball traveled on the interval $[0, 1] .$ The area under the velocity curve on $[0, 1]$ is the area of the triangle with height 32 (ft/sec) and base 1 (second), which is $A = \frac{1}{2} \cdot 1 \cdot 32 = 16$ feet. These two results are identical, in part due to the fact that we are using two different perspectives to compute the same quantity, which is distance traveled.
Observe that $s (2) - s (0) = (32 - 32) - (0 - 0) = 0 .$ This means that the ball has zero change in position on the interval $[0, 2] .$ But we already established that on the interval $[0, 1],$ the ball traveled 16 feet vertically; since the velocity becomes negative on the interval $1 < t < 2,$ there we know the ball’s position is decreasing, so it is falling back to earth. The resulting zero change in position means that at $t = 2$ the ball has returned to the location from which it was tossed. If we view the area between the velocity function and the $t$ -axis as being negative wherever $v$ is negative, then we see that the areas of the two triangles involved are opposites, which in some sense results in the “total area” being zero, matching the change in position.

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Subsection 4.1.3 When velocity is negative

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The assumption that velocity is positive on a given interval guarantees that the movement of an object is always in a single direction, and hence ensures that its change in position is the same as the distance it travels. As we saw in Example 4.13, there are natural settings in which an object’s velocity is negative, and we would like to understand this scenario as well.

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Consider a simple example where a woman goes for a walk on the beach along a stretch of very straight shoreline that runs east-west. We assume that her initial position is

s (0) = 0,

and that her position function increases as she moves east from her starting location. For instance,

s = 1

mile represents one mile east of the start location, while

s = - 1

tells us she is one mile west of where she began walking on the beach.

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Now suppose she walks in the following manner. From the outset at

t = 0,

she walks due east at a constant rate of

3

mph for 1.5 hours. After 1.5 hours, she stops abruptly and begins walking due west at a constant rate of

4

mph and does so for 0.5 hours. Then, after another abrupt stop and start, she resumes walking at a constant rate of

3

mph to the east for one more hour. What is the total distance she traveled on the time interval from

t = 0

t = 3 ?

What the total change in her position over that time?

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These questions are possible to answer without calculus because the velocity is constant on each interval. From

t = 0

t = 1.5,

she traveled

D_{[0, 1.5]} = 3 miles per hour \cdot 1.5 hours = 4.5 miles .

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t = 1.5

t = 2,

the distance traveled is

D_{[1.5, 2]} = 4 miles per hour \cdot 0.5 hours = 2 miles .

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Finally, in the last hour she walked

D_{[2, 3]} = 3 miles per hour \cdot 1 hours = 3 miles,

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so the total distance she traveled is

D = D_{[0, 1.5]} + D_{[1.5, 2]} + D_{[2, 3]} = 4.5 + 2 + 3 = 9.5 miles .

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Since the velocity for

1.5 < t < 2

v = - 4,

indicating motion in the westward direction, the woman first walked 4.5 miles east, then 2 miles west, followed by 3 more miles east. Thus, the total change in her position is

change in position = 4.5 - 2 + 3 = 5.5 miles .

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We have been able to answer these questions fairly easily, and if we think about the problem graphically, we can generalize our solution to the more complicated setting when velocity is not constant, and possibly negative.

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Figure 4.15. At left, the velocity function of the person walking; at right, the corresponding position function.

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In Figure 4.15, we see how the distances we computed can be viewed as areas:

A_{1} = 4.5

comes from multiplyimg rate times time (

3 \cdot 1.5

), as do

A_{2}

and

A_{3} .

But while

A_{2}

is an area (and is therefore positive), because the velocity function is negative for

1.5 < t < 2,

this area has a negative sign associated with it. The negative area distinguishes between distance traveled and change in position.

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The distance traveled is the sum of the areas,

D = A_{1} + A_{2} + A_{3} = 4.5 + 2 + 3 = 9.5 miles .

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But the change in position has to account for travel in the negative direction. An area above the

t

-axis is considered positive because it represents distance traveled in the positive direction, while one below the

t

-axis is viewed as negative because it represents travel in the negative direction. Thus, the change in the woman’s position is

s (3) - s (0) = (+ 4.5) + (- 2) + (+ 3) = 5.5 miles .

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In other words, the woman walks 4.5 miles in the positive direction, followed by two miles in the negative direction, and then 3 more miles in the positive direction.

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Negative velocity is also seen in the graph of the position function

y = s (t) .

Its slope is negative (specifically,

- 4

) on the interval

1.5 < t < 2

because the velocity is

- 4

on that interval. The negative slope shows the position function is decreasing because the woman is walking east, rather than west.

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To summarize, we see that if velocity is sometimes negative, a moving object’s change in position different from its distance traveled. If we compute separately the distance traveled on each interval where velocity is positive or negative, we can calculate either the total distance traveled or the total change in position. We assign a negative value to distances traveled in the negative direction when we calculate change in position, but a positive value when we calculate the total distance traveled.

Example 4.16.

Suppose that an object moving along a straight line path has its velocity

v

(in meters per second) at time

t

(in seconds) given by the piecewise linear function whose graph is pictured at left in Figure 4.17. We view movement to the right as being in the positive direction (with positive velocity), while movement to the left is in the negative direction.

Figure 4.17. The velocity function of a moving object.

Suppose further that the object’s initial position at time

t = 0

s (0) = 1 .

Determine the total distance traveled and the total change in position on the time interval $0 \leq t \leq 2 .$ What is the object’s position at $t = 2 ?$
On what time intervals is the moving object’s position function increasing? Why? On what intervals is the object’s position decreasing? Why?
What is the object’s position at $t = 8 ?$ How many total meters has it traveled to get to this point (including distance in both directions)? Is this different from the object’s total change in position on $t = 0$ to $t = 8 ?$
Find the exact position of the object at $t = 1, 2, 3, \dots, 8$ and use this data to sketch an accurate graph of $y = s (t)$ on the axes provided at right in Figure 4.17. How can you use the provided information about $y = v (t)$ to determine the concavity of $s$ on each relevant interval?

Hint.

Find the area of each triangular region formed between $y = v (t)$ and the $t$ -axis.
Recall that $v = s^{'},$ and here we are given complete information about $v .$
Be careful to address whether $v$ is positive or negative when calculating areas and adding the results.
Consider finding the area bounded by $y = v (t)$ and the $t$ -axis on each interval $[0, 1],$ $[1, 2],$ $\dots .$

Answer.

Total distance traveled is $2;$ change in position is $0 .$
$0 < t < 1$ and $4 < t < 8 .$
$s (8) - s (0) = 1 - 4 + 8 = 5 m .$
See the figure below.

Solution.

By finding the area of the triangular regions formed between $y = v (t)$ and the $t$ -axis on $[0, 1]$ and $[1, 2]$ (each of which is $1$ ), it follows that the object’s total distance traveled is $2,$ while its change in position is $0 .$ The latter is true since the net signed area bounded by $v$ on $[0, 2]$ is $1 - 1 = 0 .$
The object’s position is increasing wherever its velocity is positive, hence for $0 \leq t < 1$ and $4 < t < 8 .$
By calculating the area bounded by the curve, we find 1 unit of area on $[0, 1],$ 4 units of area on $[1, 4],$ and 8 units of area on $[4, 8],$ thus the total distance traveled on $0 \leq t \leq 8$ is $D = 1 + 4 + 8$ meters. As the change in position is given by the net signed area on this interval, we find that the change in position is

$s (8) - s (0) = 1 - 4 + 8 = 5 m .$
In the figure below, at left we list all of the areas bounded by $v$ on each one-unit subinterval. Along with the given starting point that $s (0) = 1,$ we use the resulting changes in position to plot points for the function $s .$ For instance, we know $s (1) - s (0) = 1,$ hence $s (1) = 2 .$ Similarly, $s (2) - s (1) = - 1,$ thus $s (2) = 1 .$ Continuing across the interval, we generate the function $s$ that is pictured at right. Note that the portion of $s$ from $t = 2$ to $t = 3$ is linear because $v$ is constant there, while the other parts of $s$ appear to be quadratic, as they correspond to intervals where $v$ is linear.

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Subsection 4.1.4 Left and Right Hand Sums

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At this point we have used rectangles to estimate the total area underneath a curve. We will now explore a couple of different ways of constructing these rectangles. Consider the figures below. In Figure 4.18 the rectangles are formed by setting the height of the rectangle so that the left side of the rectangle touches the curve. Conversely, in Figure 4.19 the right side of the rectangle touches the Cuve. These are commonly called left hand approximations and right hand approximations respectively.

Figure 4.18. Left hand sum approximate to the area under the graph of the equation $y = x .$

Figure 4.19. Right hand sum approximate to the area under the graph of the equation $y = x .$

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In the figure below you might notice that the left-hand approximation gives an underestimate for the total area of the curve. You might wonder what characteristics of a curve would ensure that a left-hand approximation is always underestimating the total area under the curve. You could similarly note that the right hand approximation gives an overestimate for the total area under the curve. Again, you might wonder what characteristics of a curve ensure this behavior.

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Inspection of [cross-reference to target(s) "F-4-1-Manyfigs" missing or not unique] might lead you to conclude that a left hand estimate is always an underestimate for an increasing function and an overestimate for a decreasing function, and that a right hand estimate is an overestimate for an increasing function and an underestimate for a decreasing function. This is indeed the case.

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When approximating the area under a curve the following properties hold:

A left-hand estimate will overestimate the area under any portion of the curve on which it is decreasing.
A right-hand estimate will underestimate the area under any portion of the curve on which it is decreasing.
A left-hand estimate will underestimate the area under any portion of the curve on which it is increasing.
A right-hand estimate will overestimate the area under any portion of the curve on which it is decreasing.

Example 4.20.

Consider the function whose graph is given in Figure 4.21.

On what intervals is a left-hand estimate of the area an overestimate?
On what intervals is a right-hand estimate of the area an underestimate?
On what intervals is a left-hand estimate of the area an underestimate?
On what intervals is a right-hand estimate of the area an overestimate?

Figure 4.21. A function that is increasing and decreasing on various intervals..

Hint.

Recall,

A left-hand estimate will overestimate the area under any portion of the curve on which it is decreasing.
A right-hand estimate will underestimate the area under any portion of the curve on which it is decreasing.
A left-hand estimate will underestimate the area under any portion of the curve on which it is increasing.
A right-hand estimate will overestimate the area under any portion of the curve on which it is increasing.

Answer.

$(1, 2) \cup (3, 4)$
$(1, 2) \cup (3, 4)$
$(0, 1) \cup (2, 3)$
$(0, 1) \cup (2, 3)$

Solution.

Recall,

A left-hand estimate will overestimate the area under any portion of the curve on which it is decreasing. Therefore, by inspecting Figure 4.21 we can see that the figure is decreasing on $(1, 2) \cup (3, 4) .$ Note, we can also observe the overestimate in Figure 4.22.
A right-hand estimate will underestimate the area under any portion of the curve on which it is decreasing. Therefore, by inspecting Figure 4.21 we can see that the figure is decreasing on $(1, 2) \cup (3, 4) .$ Note, we can also observe the underestimate in Figure 4.23.
A left-hand estimate will underestimate the area under any portion of the curve on which it is increasing. Therefore, by inspecting Figure 4.21 we can see that the figure is increasing on $(0, 1) \cup (2, 3) .$ Note, we can also observe the underestimate in Figure 4.22.
A right-hand estimate will overestimate the area under any portion of the curve on which it is increasing. Therefore, by inspecting Figure 4.21 we can see that the figure is increasing on $(0, 1) \cup (2, 3) .$ Note, we can also observe the overestimate in Figure 4.23.

Figure 4.22. Left hand sum approximate to the area under the graph from Figure 4.21.

Figure 4.23. Right hand sum approximate to the area under the graph from Figure 4.21.

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Subsection 4.1.5 Summary

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If we know the velocity of a moving body at every point in a given interval and the velocity is positive throughout, we can estimate the object’s distance traveled and in some circumstances determine this value exactly.
In particular, when velocity is positive on an interval, we can find the total distance traveled by finding the area under the velocity curve and above the $t$ -axis on the given time interval. We may only be able to estimate this area, depending on the shape of the velocity curve.
An antiderivative of a function $f$ is a new function $F$ whose derivative is $f .$ That is, $F$ is an antiderivative of $f$ provided that $F^{'} = f .$ In the context of velocity and position, if we know a velocity function $v,$ an antiderivative of $v$ is a position function $s$ that satisfies $s^{'} = v .$ If $v$ is positive on a given interval, say $[a, b],$ then the change in position, $s (b) - s (a),$ measures the distance the moving object traveled on $[a, b] .$
If its velocity is sometimes negative, a moving object is sometimes traveling in the opposite direction or backtracking. To determine distance traveled, we have to compute the distance separately on intervals where velocity is positive or negative, and account for the change in position on each such interval.

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Exercises 4.1.6 Exercises

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1. Estimating distance traveled from velocity data.

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A car comes to a stop six seconds after the driver applies the brakes. While the brakes are on, the following velocities are recorded:

Time since brakes applied (sec)	0	2	4	6
Velocity (ft/s)	97	49	18	0

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Give lower and upper estimates (using all of the available data) for the distance the car traveled after the brakes were applied.

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lower:

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upper:

🔗

(for each, include units

⁹

/webwork2_files/helpFiles/Units.html

)

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On a sketch of velocity against time, show the lower and upper estimates you found above..

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2. Distance from a linear velocity function.

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The velocity of a car is

f (t) = 15 t

meters/second. Use a graph of

f (t)

to find the exact distance traveled by the car, in meters, from

t = 0

t = 10

seconds.

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distance = (include units

¹⁰

/webwork2_files/helpFiles/Units.html

)

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3. Change in position from a linear velocity function.

🔗

The velocity of a particle moving along the

x

-axis is given by

f (t) = 6 - 2 t

cm/sec. Use a graph of

f (t)

to find the exact change in position of the particle from time

t = 0

t = 4

seconds.

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change in position = (include units

¹¹

/webwork2_files/helpFiles/Units.html

)

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4. Comparing distance traveled from velocity graphs.

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Two cars start at the same time and travel in the same direction along a straight road. The figure below gives the velocity,

v

(in km/hr), of each car as a function of time (in hr).

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The velocity of car A is given by the dashed, red curve, and the velocity of car B by solid, blue curve.

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(a)

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Which car attains the larger maximum velocity?

(b)

Which stops first?

(c)

Which travels farther?

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5. Finding average acceleration from velocity data.

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Suppose that an accelerating car goes from 0 mph to 57.3 mph in five seconds. Its velocity is given in the following table, converted from miles per hour to feet per second, so that all time measurements are in seconds. (Note: 1 mph is 22/15 feet per sec = 22/15 ft/s.) Find the average acceleration of the car over each of the first two seconds.

$t$	0	1	2	3	4	5
$v (t)$	0.00	28.64	49.64	64.91	76.36	84.00

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average acceleration over the first second = (include units

¹²

/webwork2_files/helpFiles/Units.html

)

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average aceleration over the second second = (include units

¹³

/webwork2_files/helpFiles/Units.html

)

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6. Change in position from a quadratic velocity function.

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The velocity function is

v (t) = t^{2} - 3 t + 2

for a particle moving along a line. Find the displacement (net distance covered) of the particle during the time interval

[- 1, 6] .

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displacement =

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7. A piecewise velocity function.

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Along the eastern shore of Lake Michigan from Lake Macatawa (near Holland) to Grand Haven, there is a bike bath that runs almost directly north-south. For the purposes of this problem, assume the road is completely straight, and that the function

s (t)

tracks the position of the biker along this path in miles north of Pigeon Lake, which lies roughly halfway between the ends of the bike path.

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Suppose that the biker’s velocity function is given by the graph in Figure 4.24 on the time interval

0 \leq t \leq 4

(where

t

is measured in hours), and that

s (0) = 1 .

Figure 4.24. The graph on the left is the biker’s velocity, $y = v (t) .$ The axes on the right will be used to plot an approximate sketch of $y = s (t) .$

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Approximately how far north of Pigeon Lake was the cyclist when she was the greatest distance away from Pigeon Lake? At what time did this occur?
What is the cyclist’s total change in position on the time interval $0 \leq t \leq 2 ?$ At $t = 2,$ was she north or south of Pigeon Lake?
What is the total distance the biker traveled on $0 \leq t \leq 4 ?$ At the end of the ride, how close was she to the point at which she started?
Sketch an approximate graph of $y = s (t),$ the position function of the cyclist, on the interval $0 \leq t \leq 4 .$ Label at least four important points on the graph of $s .$

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8. Physical interpretations of velocity.

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A toy rocket is launched vertically from the ground on a day with no wind. The rocket’s vertical velocity at time

t

(in seconds) is given by

v (t) = 500 - 32 t

feet/sec.

At what time after the rocket is launched does the rocket’s velocity equal zero? Call this time value $a .$ What happens to the rocket at $t = a ?$
Find the value of the total area enclosed by $y = v (t)$ and the $t$ -axis on the interval $0 \leq t \leq a .$ What does this area represent in terms of the physical setting of the problem?
Find an antiderivative $s$ of the function $v .$ That is, find a function $s$ such that $s^{'} (t) = v (t) .$
Compute the value of $s (a) - s (0) .$ What does this number represent in terms of the physical setting of the problem?
Compute $s (5) - s (1) .$ What does this number tell you about the rocket’s flight?

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9. Physical interpretations of velocity.

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An object moving along a horizontal axis has its instantaneous velocity at time

t

in seconds given by the function

v

pictured in Figure 4.25, where

v

is measured in feet/sec. Assume that the curves that make up the parts of the graph of

y = v (t)

are either portions of straight lines or portions of circles.

Figure 4.25. The graph of $y = v (t),$ the velocity function of a moving object.

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Determine the exact total distance the object traveled on $0 \leq t \leq 2 .$
What is the value and meaning of $s (5) - s (2),$ where $y = s (t)$ is the position function of the moving object?
On which time interval did the object travel the greatest distance: $[0, 2],$ $[2, 4],$ or $[5, 7] ?$
On which time interval(s) is the position function $s$ increasing? At which point(s) does $s$ achieve a relative maximum?

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10. Pollution Data.

🔗

Filters at a water treatment plant become dirtier over time and thus become less effective; they are replaced every 30 days. During one 30-day period, the rate at which pollution passes through the filters into a nearby lake (in units of particulate matter per day) is measured every 6 days and is given in the following table. The time

t

is measured in days since the filters were replaced.

Table 4.26. Pollution data for the water filters.

Day, $t$	$0$	$6$	$12$	$18$	$24$	$30$
Rate of pollution in units per day, $p (t)$	$7$	$8$	$10$	$13$	$18$	$35$

🔗

Plot the given data on a set of axes with time on the horizontal axis and the rate of pollution on the vertical axis.
Explain why the amount of pollution that entered the lake during this 30-day period would be given exactly by the area bounded by $y = p (t)$ and the $t$ -axis on the time interval $[0, 30] .$
Estimate the total amount of pollution entering the lake during this 30-day period. Carefully explain how you determined your estimate.

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