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Determining distance traveled from velocity

In the first section 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.

Introduction

In the first section 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, \[\begin{aligned}\end{aligned}\].

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

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.

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 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?

Exploration
Exploration

Area under the graph of the velocity function

In Preview Activity, 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.

The left-hand graph of Figure 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 \(y = v(t)\) on \([1,1.5]\) is \[\begin{aligned}\end{aligned}\].

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 \[\begin{aligned}\end{aligned}\], where \(\Delta t\) 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 this interactive.

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 graph at right in Figure 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, \[\begin{aligned}\end{aligned}\].

Note that because \(v\) is decreasing on \([1,1.5]\), \(A_2 = 1.25\) is an over-estimate of the actual distance traveled.

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, 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?

Condensed — the full section is in Boelkins, Active Calculus.

Two approaches: area and antidifferentiation

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.

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.

In Activity, 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.

For a simple example, consider the situation from Preview Activity, where a person is walking along a straight line with velocity function \(v(t) = 3\) mph.

On the left-hand graph of the velocity function in Figure, we see the relationship between area and distance traveled, \[\begin{aligned}\end{aligned}\].

In addition, we observe 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 \(s(t) = 3t\), then \(s'(t) = 3\), so \(s(t) = 3t\) 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 \[\begin{aligned}\end{aligned}\].

The quantity \(s(1.5) - s(0.25)\) is the person's change in position on \([0.25,1.5]\), as well as the distance traveled since their velocity is always positive. In this example there are profound ideas and connections that we will study throughout Chapter.

Condensed — the full section is in Boelkins, Active Calculus.

When velocity is negative

The assumption that the 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 Activity, there are natural settings in which an object's velocity is negative, and we would like to understand this scenario as well.

Consider a simple example where a person 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.

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 \(t = 0\) to \(t = 3\)? What is the total change in her position over that time?

These questions are possible to answer without calculus because the velocity is constant on each interval. From \(t = 0\) to \(t = 1.5\), she traveled \[\begin{aligned}\end{aligned}\].

On \(t = 1.5\) to \(t = 2\), the distance traveled is \[\begin{aligned}\end{aligned}\].

Finally, in the last hour she walked \[\begin{aligned}\end{aligned}\], so the total distance she traveled is \[\begin{aligned}\end{aligned}\].

Since the velocity for \(1.5 \lt t \lt 2\) is \(v = -4\), indicating motion in the westward direction, the person first walked 4.5 miles east, then 2 miles west, followed by 3 more miles east. Thus, the total change in her position is \[\begin{aligned}\end{aligned}\].

It's also valuable to think about our conclusions from a graphical perspective.

Condensed — the full section is in Boelkins, Active Calculus.

Summary

  • 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.

Practice (5)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. A rainfall rate of 0.30 inches or more per hour is considered heavy rain, according to the American Meteorological Society.

    Suppose that a heavy thunderstorm lasts for 10 minutes and during those 10 minutes rain falls at a rate of 0.30 inches per hour.

    1. What do you get if you multiply \(0.30 \dfrac{\text{ inches}}{\text{hour}} \cdot \dfrac16 \text{hour}\)? Include units.

    2. What does your answer mean in the context of the thunderstorm?

  2. 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 \(v(t)= 500-32t\) feet/sec.

    1. 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 \(t = a\)?

    2. Find the value of the total area enclosed by \(y = v(t)\) and the \(t\)-axis on the interval \(0 \le t \le a\). What does this area represent in terms of the physical setting of the problem?

    3. Find an antiderivative \(s\) of the function \(v\). That is, find a function \(s\) such that \(s'(t) = v(t)\).

    4. Compute the value of \(s(a) - s(0)\). What does this number represent in terms of the physical setting of the problem?

    5. Compute \(s(5) - s(1)\). What does this number tell you about the rocket's flight?

    Zbulo përgjigjen

    1. Setting \(v(t) = 0\), we find \(500 - 32t = 0\), and thus \(t = \frac{500}{32} = \frac{125}{8} = 15.625\). This is the time that the rocket's velocity changes from positive to negative, and thus the point when the rocket's position changes from increasing to decreasing: it's when the rocket reaches its maximum height.

    2. Since \(v\) is a positive linear function that connects the points \((0,500)\) and \((15.625,0)\) on the interval \([0, 15.625]\), the region enclosed by the velocity function and the \(t\)-axis is a right triangle with vertical leg of length \(500\) and horizontal leg of length \(15.625\). The area of this triangle is \(A = \frac{1}{2}bh = \frac{1}{2}(15.625)(500) = 3906.25\). Because this is the area under the velocity curve (where velocity is positive) this area represents the vertical distance traveled by the rocket on the time interval \([0, 15.625]\).

    3. One function that has derivative \(v(t) = s'(t) = 500-32t\) is \(s(t) = 500t - 16t^2\).

    4. Using our result in (c), \(s(15.625) - s(0) = 500(15.625) - 16(15.625)^2 - (0 - 0) = 3906.25\). Thus, the change of the rocket's position on \([0,15.625]\) is \(3906.25\) feet, which is also the distance the rocket traveled vertically on the time interval, since velocity is positive throughout.

    5. We find that \(s(5) - s(1) = 1616\), which tells us that the rocket rose \(1616\) feet on the \(4\)-second interval from \(t = 1\) to \(t = 5\).

  3. An object moving along a horizontal axis has its instantaneous velocity at time \(t\) in seconds given by the function \(v\) pictured in Figure, 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.

    1. Determine the exact total distance the object traveled on \(0 \le t \le 2\).

    2. What is the value and meaning of \(s(5) - s(2)\), where \(y = s(t)\) is the position function of the moving object?

    3. On which time interval did the object travel the greatest distance: \([0,2]\), \([2,4]\), or \([5,7]\)?

    4. On which time interval(s) is the position function \(s\) increasing? At which point(s) does \(s\) achieve a relative maximum?

    Zbulo përgjigjen

    1. To find the exact total distance traveled, we need to calculate the area under the velocity function on the interval \([0,2]\). This area is the sum of the areas of the triangular region of height \(1\) and base length \(1\) on the interval \([0,1]\) plus the area of the quarter circle of radius \(1\) on the interval \([1,2]\). So the total distance the object traveled on \(0 \le t \le 2\) is \[\begin{aligned}\end{aligned}\].

    2. The meaning of \(s(5)-s(2)\) is the change in position, or total displacement, of the object on the time interval \([2,5]\). Note that \(s(2)\) was found to be \(\frac{1}{2} + \frac{1}{4} \pi\) in part (a) and \[\begin{aligned}\end{aligned}\]. So, \(s(5) - s(2) = -2\), which means that the object moved 2 feet back toward its original position from time \(t=2\) to time \(t=5\).

    3. In part (a) we calculated the distance traveled by the object on the interval \([0,2\) to be approximately \(1.285\) feet. On the time interval \([2,4]\), the object is moving in the opposite direction from the motion on the interval \([0,2\), and the total distance traveled is the area \(v(t)\) bounds below the \(t\)-axis on the interval \([2,4\). This area is the sum of the areas of the triangle of height \(1\) and base length \(1\) on the interval \([2,3]\) and the area of the rectangle of height \(1\) and length \(1\) on the interval \([3,4\). So the total distance traveled by the object on the interval \([2,4]\) is \[\begin{aligned}\end{aligned}\]. The total distance traveled by the object on the interval \([5,7\) is the area \(v(t)\) bounds above the \(t\)-axis on the interval \([5,7]\). This area is the area of the semicircle of radius \(1\)on the interval \([5,7]\). So the total distance traveled by the object on the interval \([5,7]\) is \[\begin{aligned}\end{aligned}\]. Thus, the object traveled the greatest distance on the time interval \([5,7]\).

    4. We know that the position function is increasing when the velocity is positive, so \(s\) is increasing on the intervals \((0,2)\) and \((5,7)\). Since \(v(t)\) is positive to the left of \(t=2\) and then negative to the right of \(t=2\), the position function \(s\) is increasing to the left of \(t=2\) then decreasing to the right of \(t=2\). Thus, the position function has a relative maximum point at \(t=2\). There are no other points with this property, so this is the only relative maximum value of \(s\) on this interval. We also can note that \(s\) increases from \(t=5\) to \(t=7\), culminating in a local maximum value at \(t=7\).

  4. 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 \(v(t)= 500-32t\) feet/sec.

    1. 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 \(t = a\)?

    2. Find the value of the total area enclosed by \(y = v(t)\) and the \(t\)-axis on the interval \(0 \le t \le a\). What does this area represent in terms of the physical setting of the problem?

    3. Find an antiderivative \(s\) of the function \(v\). That is, find a function \(s\) such that \(s'(t) = v(t)\).

    4. Compute the value of \(s(a) - s(0)\). What does this number represent in terms of the physical setting of the problem?

    5. Compute \(s(5) - s(1)\). What does this number tell you about the rocket's flight?

    Zbulo përgjigjen

    1. Setting \(v(t) = 0\), we find \(500 - 32t = 0\), and thus \(t = \frac{500}{32} = \frac{125}{8} = 15.625\). This is the time that the rocket's velocity changes from positive to negative, and thus the point when the rocket's position changes from increasing to decreasing: it's when the rocket reaches its maximum height.

    2. Since \(v\) is a positive linear function that connects the points \((0,500)\) and \((15.625,0)\) on the interval \([0, 15.625]\), the region enclosed by the velocity function and the \(t\)-axis is a right triangle with vertical leg of length \(500\) and horizontal leg of length \(15.625\). The area of this triangle is \(A = \frac{1}{2}bh = \frac{1}{2}(15.625)(500) = 3906.25\). Because this is the area under the velocity curve (where velocity is positive) this area represents the vertical distance traveled by the rocket on the time interval \([0, 15.625]\).

    3. One function that has derivative \(v(t) = s'(t) = 500-32t\) is \(s(t) = 500t - 16t^2\).

    4. Using our result in (c), \(s(15.625) - s(0) = 500(15.625) - 16(15.625)^2 - (0 - 0) = 3906.25\). Thus, the change of the rocket's position on \([0,15.625]\) is \(3906.25\) feet, which is also the distance the rocket traveled vertically on the time interval, since velocity is positive throughout.

    5. We find that \(s(5) - s(1) = 1616\), which tells us that the rocket rose \(1616\) feet on the \(4\)-second interval from \(t = 1\) to \(t = 5\).

  5. An object moving along a horizontal axis has its instantaneous velocity at time \(t\) in seconds given by the function \(v\) pictured in Figure, 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.

    1. Determine the exact total distance the object traveled on \(0 \le t \le 2\).

    2. What is the value and meaning of \(s(5) - s(2)\), where \(y = s(t)\) is the position function of the moving object?

    3. On which time interval did the object travel the greatest distance: \([0,2]\), \([2,4]\), or \([5,7]\)?

    4. On which time interval(s) is the position function \(s\) increasing? At which point(s) does \(s\) achieve a relative maximum?

    Zbulo përgjigjen

    1. To find the exact total distance traveled, we need to calculate the area under the velocity function on the interval \([0,2]\). This area is the sum of the areas of the triangular region of height \(1\) and base length \(1\) on the interval \([0,1]\) plus the area of the quarter circle of radius \(1\) on the interval \([1,2]\). So the total distance the object traveled on \(0 \le t \le 2\) is \[\begin{aligned}\end{aligned}\].

    2. The meaning of \(s(5)-s(2)\) is the change in position, or total displacement, of the object on the time interval \([2,5]\). Note that \(s(2)\) was found to be \(\frac{1}{2} + \frac{1}{4} \pi\) in part (a) and \[\begin{aligned}\end{aligned}\]. So, \(s(5) - s(2) = -2\), which means that the object moved 2 feet back toward its original position from time \(t=2\) to time \(t=5\).

    3. In part (a) we calculated the distance traveled by the object on the interval \([0,2\) to be approximately \(1.285\) feet. On the time interval \([2,4]\), the object is moving in the opposite direction from the motion on the interval \([0,2\), and the total distance traveled is the area \(v(t)\) bounds below the \(t\)-axis on the interval \([2,4\). This area is the sum of the areas of the triangle of height \(1\) and base length \(1\) on the interval \([2,3]\) and the area of the rectangle of height \(1\) and length \(1\) on the interval \([3,4\). So the total distance traveled by the object on the interval \([2,4]\) is \[\begin{aligned}\end{aligned}\]. The total distance traveled by the object on the interval \([5,7\) is the area \(v(t)\) bounds above the \(t\)-axis on the interval \([5,7]\). This area is the area of the semicircle of radius \(1\)on the interval \([5,7]\). So the total distance traveled by the object on the interval \([5,7]\) is \[\begin{aligned}\end{aligned}\]. Thus, the object traveled the greatest distance on the time interval \([5,7]\).

    4. We know that the position function is increasing when the velocity is positive, so \(s\) is increasing on the intervals \((0,2)\) and \((5,7)\). Since \(v(t)\) is positive to the left of \(t=2\) and then negative to the right of \(t=2\), the position function \(s\) is increasing to the left of \(t=2\) then decreasing to the right of \(t=2\). Thus, the position function has a relative maximum point at \(t=2\). There are no other points with this property, so this is the only relative maximum value of \(s\) on this interval. We also can note that \(s\) increases from \(t=5\) to \(t=7\), culminating in a local maximum value at \(t=7\).

Symbols used here

\approx
approximately equal
Equal to the precision shown, not exactly.
\leq,\ \geq
less/greater than or equal
Inequalities that allow equality; < and > exclude it.
\pi
pi
Ratio of a circle's circumference to its diameter, 3.14159…
e
Euler's number
2.71828…, the base whose exponential is its own derivative.
\infty
infinity
Not a number: "grows without bound" in limits and intervals.
\sin,\ \cos,\ \tan
sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.
\log_b x,\ \ln x
logarithm, natural log
The exponent b must be raised to for x; ln uses base e.
\sum_{k=1}^{n} a_k
summation
Add a_k for k = 1 up to n.
\lim_{x \to a} f(x)
limit
The value f(x) approaches as x approaches a.
f'(x),\ \frac{dy}{dx}
derivative
Instantaneous rate of change; slope of the graph.
\int f(x)\,dx,\ \int_a^b
integral
Antiderivative (indefinite) or signed area from a to b (definite).
y',\ y''
first and second derivative of y
Prime notation for derivatives with respect to x (or t).
C,\ C_1,\ C_2
arbitrary constants
Constants of integration fixed by initial conditions.

How to: Determining distance traveled from velocity

  1. 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 the time interval?
  2. How is the problem of finding distance traveled related to finding the area under a certain curve?
  3. What does it mean to antidifferentiate a function and why is this process relevant to finding distance traveled?
  4. If velocity is negative, how does this impact the problem of finding distance traveled?

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

Provo timen.

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

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