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. 

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.] 

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

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

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

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

• 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.

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.

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.
16. 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.
17. 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:

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.