IV.C. Motion and Differential Equations.
Preface: The interpretation of the derivative as a rate makes the study of differential equations a significant part of the study of rates in many contexts. In this section we consider the two rates traditionally associated with a moving object, namely the velocity and the acceleration of the object. Using the very simple techniques of the preceding sections together with some elementary physical properties of freely falling objects, we can begin to see the scientific power of being able to solve differential equations. Beyond the applications of this section, the motion interpretation is important because it suggests results that extend to many other contexts.
 Historical Note: Nicholas Oresme (1323-1382) was perhaps the first person to fully understand and explain that the uniform acceleration of a moving object would lead to a parabolic motion. His work on motion also made a clear connection between the motion of an object and a geometric analysis of what we would describe today as a the graphical representation of a moving object. Over a hundred years later, in about 1638, Galileo Galilei (1564 - 1642) observed through experimentation that an object either falling or rolling without any external interference would move with a uniform acceleration. From this, independent of the work of Oresme, he came to the same conclusions and described the motion of a cannon ball as following a parabolic path during its flight. Still later in history Sir Isaac Newton (1643-1727) provided a general theory of gravitation that explained the uniform acceleration of objects both small (like apples) or large (like planets) and thereby completed a mathematical model for the motion of falling objects.
Constant Velocity
We'll start by analyzing a  simple problem where an object is moving with a constant velocity.

Example IV.C.1: A remote control car is 30 feet from the wall and is travelling at a constant rate of 2 feet/second toward the wall. When will the car reach the wall?
Solution: The car will travel 2 feet for each second, so in 15 seconds it will cover 2 . 15 = 30 feet, the required distance to the wall.

But what does this have to do with differential equations?

Discussion: First, we are interested in the position of the car as a function of time. We suppose the car is moving on a straight line toward the wall at a constant rate. Let's use S(t) for the measure in feet of the distance travelled by the car after t seconds, so S(0) = 0.. The assumption about the rate at which the car is travelling translates into an equation about the velocity of the car at time t, namely, v(t) = S'(t) = 2. The car will have travelled 30 feet when it reaches the wall. Our question is for what value of t does S(t) = 30?

We've assumed that S satisfies the differential equation S'(t) = 2 with initial condition S(0) = 0. v(t) = S'(t) = 2; S(t) = 2t + C. S(0) = 0; S(t) = 2t; S(15) = 30

The general solution to this differential equation is S(t) = 2t + C.

The initial condition that S(0) = 0 implies C = 0, so the particular solution  is S(t) = 2t.

Now the problem is to find the value of t when S(t) = 30.
So we set 2t = 30 and solve for t, giving t = 15 as was suggested originally.

[This certainly is a long way to go for such a simple result. The example merely illustrates how differential equations might be used for such a simple problem.]

Constant Acceleration
Now we should be ready to tackle a slightly more difficult problem from physics. According to Newton's laws of gravity and motion, the acceleration of a moving object (like a cannon ball) influenced only by the gravitational force of the earth is a constant. This constant can be estimated by physical experiments and is traditionally denoted g. We use g = 32 ft/sec 2 or 9.8m/sec2 for our calculations. The sign of g is determined by our point of view. If we measure the distance the object has travelled down from a starting position above the earth (say dropped from the top of a tall tower or cliff) then g is a positive number. On the other hand, if we measure the object's distance above the earth, then g is a negative number. To make this more concrete, here are two examples similar to Example IV.C.1.

Example IV.C.2: An anvil is dropped from the top of a cliff that is 144 feet about ground level. When will the anvil reach ground level? [Before reading further, you might try to estimate an answer.]

Solution: First, we are interested in the position of the anvil as a function of time and we suppose the anvil is moving on a straight line toward the ground. According to Newton's laws the acceleration of the anvil is a constant rate. We'll let S(t) be the measure in feet of the distance travelled by the anvil t seconds after the anvil is dropped from the top of the tower, so S(0) = 0. Now the physical assumption about the rate at which the anvil is travelling translates into an equation about the acceleration of the anvil at time t, namely, a(t) = v'(t) = S''(t) = 32(ft/sec 2).

Since the anvil was dropped, rather than thrown, its initial velocity is assumed to be 0 ft/sec, so we also have v(0)= S'(0) = 0.

The problem is now to find the value of t when S(t) = 144 where S satisfies the differential equation S''(t) = 32 with initial conditions S(0) = 0 and S'(0) = v(0) = 0.

First we'll find how the velocity is changing as a function of time.
Since S''(t)= v'(t) = 32, we see that v(t) = S'(t) = 32t + C 1.
Using v(0) = 0 we have C 1 = 0 and thus v(t) = S'(t) = 32t.

Now we can find the position function.
Since S'(t) = v(t) = 32t, we have that S(t) = 16t 2 + C 2.
Since S(0) = 0, we have C 2 = 0 and thus S(t) = 16t 2 .

So our problem is now to find t when 16t 2 = 144.

Solving this gives t 2=9, so t = 3 OR t = -3.
For our problem only a positive value of t makes sense, so we have found that the anvil will reach the ground 3 seconds after it is dropped from the tower.

We'll end this section with an example of an object thrown from ground level to illustrate how gravity can have a negative value from a ground level point of view. (This is the point of view adopted by most people.)

Example IV.C.3. A ball is projected vertically from the ground into the air by a throwing machine and reaches its greatest height after exactly three seconds. Find the initial velocity of the ball and the greatest height it reached .
Solution: Let S(t) be the height of the ball above ground level at time t seconds after it was thrown. We are given that S(0) = 0. We are also told that ball reaches its maximum height after 3 seconds, which means from the calculus theory of extremes that 3 is a critical point for the position function, so S'(3) = 0.

In this case, since the velocity of the ball is decreasing we use a negative acceleration for gravity, S''(t) = a(t) = -32 ft/sec.
So S'(t) = v(t) = -32t + C1 for some constant C1 and we have observed that when t =3  we have a critical point, so  0 = S'(3)= -32(3) + C1. Solving for C1 , we find that C1 = 96 ft/sec and

S'(t) = -32t + 96. (*)

Using this equation we have now that the initial velocity of the ball was S'(0) = 96 ft/ sec. We still need to find the position of the ball at time 3 seconds (when the ball was at its greatest height). By solving the differential equation (*), we find that

S(t) = -16t 2 + 96t + C2. (**)

Since the ball was thrown from ground level, we have that S(0) = 0, so C 2 = 0 and

S(t) = -16t 2 + 96t. (***)

We were told that the maximum height was reached after three seconds, so the maximum height's value is given by

S(3) = -16(3) 2 + 96(3) = -144 + 288 = 144ft
• It is a remarkable feature of physics and the calculus that by assuming Newton's Laws we can know about the position and velocity of freely falling object at every instant merely from knowledge of the position and velocity of the object at one or two particular times. In fact the last example illustrates that the single observation of how long it takes for an object to reach it maximum height is sufficient to describe the motion of the object completely.
• Be sure to try exercise 7 in the problems for this section to derive the general position formula for a freely falling object. It is not something to be memorized, but is something that should be understood.
• Indefinite integrals are frequently used to express the analysis we have done here with differential equations. For example, the relation of acceleration, a, to velocity, v, might be expressed by ò a(t) dt = v(t) + C, while  ò v(t) dt = S(t) + C represents the connection of velocity, v, to position, S.

Problems IV.C.
For problems 1 - 9 find the position S(t) at time t of an object moving on a straight line from the information given about the velocity, acceleration, and position of the object. For problems 1-7, find the change in position between time 0 and time 1.
1. v(t) = 8 and s(0) = 5.
2. v(t) = 10t and s(1) = 3.
3. v(t) = 3t - 5 and s(0) = 5.
4. v(t) = 5t 2 - 3t and s(0) =5.
5. a(t) = -32, v(0) = 16 and s(0) = 8.
6. a(t) = 5t, v(0) = 10 , and s(0) = 5.
7. a(t) = g, v(0) = v 0, and s(0) = s 0 where v 0 and s 0 are constants.
8. v(t) = cos(t) and s(0) = 1.
9. a(t) = cos(t), v(0) = 1 and s(0) = 5.
10. A ball is thrown straight up from ground level and reaches its greatest height after 5 seconds. Find the initial velocity of the ball and the value of its maximum height above ground level.
11. A ball is dropped from a window 32 feet above ground level. How long will it take the ball to reach ground level? What will be the ball's velocity at the instant it reaches ground level?
12. A ball is thrown up from a window 32 feet above ground level and reaches its maximum height after 3 seconds. How long will it take the ball to reach ground level? What will be the ball's velocity at the instant it reaches ground level?
13. A ball that is thrown from one person to another will rise and fall so that its height above ground level is precisely the same as that of a ball thrown straight up with the same initial rate of motion in the vertical direction. Suppose that both people handle the ball (either throwing or catching it) at 5 feet above a level surface. The ball travels at a constant rate of 20 feet per second over the surface for 60 feet before being caught. What is the maximum height of the ball above the surface during its flight?
14. On the planet Ush a ball dropped from 100 meters above the ground takes exactly 8 seconds to reach the ground. How long will it take a ball dropped from 200 meters above the ground to reach the surface?
15. Two teammates on the USH track team were out for an afternoon run. Franky ran f (t) meters in t seconds while Johnny ran g(t) meters in the same amount of time.

16. Suppose that f '(t) < g'(t) for all t and f (0) = g(0). Explain why at any time t, Franky has run further than Johnny.
17. Suppose a small red ball is dropped 100 feet from the window of an office building while a small blue ball will be dropped 40 ft from another window in the building a little later. Describe how to drop the two balls so they will reach ground level at precisely the same moment.
18. Research Projects:
1. Look up the work of Oresme. Write an exposition explaining how he used geometry to analyze the question of the distance travelled by an object that is uniformly accelerating.
2. Look up the work of Galileo. Write a short paper explaining how he used geometry to analyze the question of the distance travelled by an freely falling object.
3. Write a short paper comparing the work of Oresme and Galileo on moving objects.
4. Find out what Newton's Laws of Motion and Gravity are. In a short paper explain the relation of these laws to the following statement:

5. For objects moving close to the surface of the earth, the acceleration due to the influence of the gravitational force of the earth is constant for most practical purposes.