© 2003 by Karl Hahn

This page is still under construction
In past units, we've already discussed plenty of math illustrations using moving trains. Recall that we were able to characterize the train's position on the track as a function of time, t. That is, we said that the train's position was x(t). As time passed, the train's position would change, which is what you expect with any object in motion. We also mentioned that the train's velocity, v(t), was also a function of time, since the train had to slow down and stop at the station, then start up again when it left the station, trundling slowly back up to speed. Most importantly, we discussed the crucial relationship between position as a function of time and velocity as a function of time. That is that velocity is the derivative with respect to time of position.
dx v(t) = x'(t) = 
The equation above gives you this same relationship in math symbols. So if you know the train's position as a function of time, it's easy enough to find its velocity, v(t), provided you remember what you learned about taking derivatives. Simply apply the rules for taking derivatives to the position function, and you've got velocity.
But suppose you only know the velocity of the train as a function of time, but not its position. If you would like to find out what the position is as a function of time, then you need to recall the fundamental theorem of calculus, which tells you that the opposite of finding a derivative is finding an integral. So by inverting equation 12.11, we get

So for example, you might know that v(t) = V_{f}(1  e^{kt}), where V_{f} and k are constants. This velocity function starts out, when t = 0 with the train stopped. As time progresses, the train accelerates with its velocity asymptotically approaching a final velocity of V_{f}. Setting up the integral for position based upon velocity being position's derivative, we have:
x(t) = V_{f} 
(1  e^{kt}) dt = V_{f} 
1 t + 
e^{kt} 
+ C eq. 12.13 
The trouble with the above is that the train's position function, x(t), is subject to an undetermined constant, C. The only way to get past that is to know what the train's position was at some particular time, t_{0}. Armed with that information, we can turn the indefinite integral above into a definite integral, and the undetermined constant, C, will go away.
Let's say that the Denver train station is our origin. Let's also say that t_{0} = 0, and at the time, t_{0}, the train is at the Denver station. Now we rewrite equation 12.13 as a definite integral as follows:
x(t) = V_{f} 
t t_{0} 
(1  e^{ks}) ds = V_{f} 
1 s + 
e^{ks} 
t s=t_{0} 
eq. 12.14a 
So what do all these funny symbols mean? Let's start at the left. You already knew that x(t) denotes where the train is on the track. Since we decided that the Denver station is at the origin (and let's say that east is the positive direction), then x(t) denotes how far east of the Denver station the train is at time, t. Note that t is the time at which we would like to know the train's position.
As for the integral, it says that we are integrating the velocity function, v(t) = V_{f}(1  e^{kt}), from time t_{0} (which we said earlier was zero) to time t. I would like you to notice, though, the change in variables inside the integral. I did not use the variable, t, inside the integral. Why not? This is a matter of form. We would like to end up with a function of t. But in a definite integral, the variable with respect to which you integrate disappears at the end. We don't want t to disappear  it is our independent variable after all. Also we don't want to use the same variable in the integral's limits as we do inside the integral. To do so would introduce a confusion factor that we can live without. For those reasons on problems like this (whenever the independent variable appears in the limits of the definite integral), we use a dummy variable inside the integral. I have chosen the symbol, s, for the dummy. It doesn't matter what you choose (as long as it's a symbol you haven't already used elsewhere in the equation) because it's going to disappear in the end anyway.
Finally we have the bracketed expression. Inside is the indefinite integral of v(t), but using the dummy variable, s, as the independent variable instead of t. The limits on the righthand bracket are saying, evaluate this bracketed function at s = t_{0} and at s = t, and then take the difference of the latter minus the former. So let's do that.
x(t) = V_{f} 
t + 
1 
e^{kt} 
 V_{f} 
t_{0} + 
1 
e^{kt0} 
eq. 12.14b 
Observe that the dummy variable, s, has already disappeared at this point. We can simplify this further by remembering that we stipulated earlier that t_{0} = 0. Plugging that in (and recalling that e^{0} = 1) we get as our final position function:
x(t) = V_{f} 
t + 
1 
e^{kt}  
1 
eq. 12.14c 
Notice that at t = 0, we have x(t) = 0, which puts the train at the station at that time. Notice also that the train does not move very much at first. In the first second or so after the train starts up, although t increases, it turns out that (1/k)e^{kt} decreases by nearly the same amount. So in the first few seconds, x(t) hardly increases at all. This is the behavior you expect from a train that is just starting up  that it does not move very much at first. The graph to the right uses V_{f} = 10 meters per second, and k = 1/6 reciprocal seconds (note that k must be in the units that are reciprocal of t so that the units cancel inside the exponential).
So what is the third trace in figure 12.11? The red trace is the derivative with respect to time of v(t). That means it is also the second derivative of x(t). In general, that function, a(t), is acceleration.

In the case of our train, a(t) is shown on the graph in units of meters per second per second, or more often expressed as meters per second^{2}. An acceleration of 1 meter per second^{2} means that with each second, velocity increases by 1 meter per second.
You can see from the graph that at time, t = 0, the engineer opens the throttle and the train immediately begins to accelerate. But as the train goes closer and closer to its top speed of 10 meters per second, the acceleration goes closer and closer to zero. You can find the equation of the acceleration as a function of time for the train by taking the derivative of the velocity function, v(t) = V_{f}(1  e^{kt}).
a(t) = V_{f} k e^{kt} eq. 12.16
This, of course, means that if you know acceleration but not velocity, then you can invert equation 12.15 to find velocity from acceleration, just as we did for finding position from velocity.

This time your velocity function, v(t), would be subject to an undetermined constant. But in the example of the train, we know that at time t = 0, not only is the train at the station, it is stopped at the station. So we know that v(0) = 0, and from that we can set up a definite integral in just the same way as we did for finding position from velocity.
One type of problem that they like to give to first year students is where a(t) is a constant. Let's take the example of an object falling under the influence of gravity with no resistance due to friction with the air. Near the surface of the earth, a falling object has a constant acceleration of a(t) = 9.8 meters per second^{2} (that's 32 feet per second^{2} in the English system of measurements). The sign is negative because objects fall down, and the conventional sense is that up is positive and down is negative. From that we get
v(t) = 9.8 
dt = 9.8 t + v_{0} eq. 12.19a 
I have chosen v_{0} as the name of the undetermined constant here, rather than C, because if you put t = 0 into the above, you find that v_{0} is how fast the object was moving at time zero. Without yet knowing what that undetermined constant, v_{0}, is, you can integrate this again to get x(t).
x(t) = 
(9.8t + v_{0}) dt =  
9.8 
t^{2} + v_{0} t + x_{0} eq. 12.19b 
This time the undetermined constant is x_{0}. If you put t = 0 into the above, you will find that x_{0} is how high the object was at time zero.
You have probably seen equation 12.19b before, or at least the expression on the right. In general, if a(t) is constant, then
1 x(t) =which you have probably also seen before. It is derived the same way as we got equation 12.19b, except instead of 9.8, just use the symbol, a. Again v_{0} indicates how fast the object was going at time, t = 0, and x_{0} indicates where the object was at time t = 0.a t^{2} + v_{0} t + x_{0} eq. 12.19c 2
Example problem: From the top of a 300 meter high cliff a ball is thrown straight up at a velocity of 20 meters per second. Assuming the ball misses the cliff on the way back down, how high is it 5 seconds after it is thrown, and how fast is it going? Solution: You have x_{0} = 300 meters and v_{0} = 20 meters per second. Because the ball is falling under the influence of gravity near the earth's surface, we use constant acceleration of a = 9.8 meters per second^{2}. Let t = 0 be the moment the ball is thrown. The problem is simply asking you what the state of thing is at t = 5 seconds.
9.8 x(5) =  
5^{2} + 20 × 5 + 300 = 277.5 meters 
Recall that the cliff is at 300 meters. The ball has already risen as high as it will go, and then fallen to 22.5 meters below the edge of the cliff. To find the velocity after 5 seconds, just apply the velocity formula:
v(5) = 9.8 × 5 + 20 = 29 meters per secondThe velocity being negative indicates that the ball is moving downward at that moment. They might also ask you at what moment is the ball at its highest? This is just a maximization problem, where you find the t that maximizes x(t). Since v(t) is the derivative of x(t), simply solve for when v(t) is zero:
0 = 9.8 t + 20 20 t =I hope what we've done so far with constant acceleration is all review to you. In high school algebra classes they usually give you the formulas I have presented here for constant acceleration and have you apply them. The only thing that is new here is that the use of calculus and integrals shows you where these formulas come from.= 2.04 seconds 9.8
Boundary Conditions: In the thrownball problem above, you knew the acceleration function, but to get from that to a position function, you had to know two additional pieces of information. These were given in the problem as how fast the ball was thrown, v_{0}, and where it was thrown from, x_{0}. These are called boundary conditions (or sometimes initial conditions). We had to integrate twice to get from acceleration to position. The number of boundary conditions necessary to arrive at a solution was also two. Do you see the pattern? The number of boundary conditions needed to find a solution is always the same as the number of times you have to integrate to find the solution. Boundary conditions are simply reference points by which you can resolve the values of the undetermined constants that arise from integration. Two integrations yield two undetermined constants. Hence two boundary conditions are needed to resolve them.
But do the boundary conditions have to be the way we saw them in the example problem? No. Neither boundary condition has to be at t = 0. Nor does one boundary condition have to be velocity and the other position. You need to know either position at two distinct times, position at one time and velocity as possibly, but not necessarily the same time. Let's take the thrown ball again. If you know that at t = 1 second the ball's position is 6 meters and at t = 2 seconds the ball's velocity is 2 meters per second, how do we find how high the ball is at t = 1.5 seconds? Just substitute the numbers:
9.8 x(1) = 6 meters =  
1^{2} + v_{0} × 1 + x_{0} 
v(2) = 2 meters per second = 9.8 × 2 + v_{0} 
Now solve for v_{0} and x_{0}. From the second equation you can easily solve to find that v_{0} = 17.6 meters per second. Now put that v_{0} into the first equation and you find that x_{0} = 6.7 meters. So one interpretation is that the ball was thrown from a 6.7 meter deep pit at time zero. Another interpretation could be that the ball was thrown from ground level, but at some time after time zero (as an exercise, see if you can solve for when that would be using the quadratic formula). If position were given at two distinct times, the problem would be similar but you would have to solve for v_{0} and x_{0} simultaneously. You would write the position equation (12.19c) twice, once for each boundary condition. So if the conditions are, for example, x(1) = 16 meters and x(2) = 22 meters, your equations would be
9.8 x(1) = 16 meters =  
1^{2} + v_{0} × 1 + x_{0} 
and
9.8 x(2) = 22 meters =  
2^{2} + v_{0} × 2 + x_{0} 
Subtracting the first equation from the second and gathering terms, you get v_{0} = 20.7 meters per second. Substitute that back into either equation and you find that x_{0} = 0.2 meters.
Let's try a more complicated example  one in which the acceleration is not constant. A rocket is fired from rest in space. The initial mass of the rocket is 110 kg. But as the rocket burns its fuel, it loses mass at a steady rate until burnout 10 seconds after firing, at which time its mass is down to 10 kg. Clearly it's burning fuel at a rate of 10 kg/second. The rocket's thrust is constant during the burn at 1000 Newtons (a Newton is the metric measurement unit of force; one Newton accelerates a 1 kilogram mass at a rate of 1 meter/second^{2}). So the equation for the rocket's mass is
m(t) = 110  10tAccording to Newton's second law of motion, the acceleration at any time is the thrust divided by the rocket's mass.
a(t) = 
1000 
v(t) = 
t 0 
a(u) du = 
t 0 
1000 du 
1 
t 0 
v(t) = 100 ln 
110 
meters/second for 0 < t < 10 seconds 
In this section you covered motion problems where an object is falling due to gravity through a vacuum. For a real challenge, have a look at the solution for an object falling through the air. This problem puts together various calculus topics that you have learned so far.