Suppose a compressed gas tank is used for thrust in space like in the Wall-E movie when the fire extinguisher is used. As the valve of the tank is opened to allow the gas to escape, the pressure drops quickly at first, but then ever slower. This is because with a high tank pressure, at first the gas is expelled very rapidly, but as that gas is lost and the tank pressure drops, the rate of expulsion also drops. [br][br]So we have a situation in which the rate of change of gas loss is proportional to the amount of gas remaining in the tank. This should sound like an exponential function, since it's the only function for which the rate of change (derivative) is always proportional to the function. [br][br][color=#1e84cc]An aside:[br]I need to mention that nature is not quite that simple. In an actual high pressure tank, the stored gas tends to be in liquid state, and upon release of gas, pressure drops inside the tank, some of the liquid evaporates, and then evaporative cooling takes place. With propane tanks or CO2 cartridges used to power BB guns or to inflate bike tires, you tend to find that the tanks get literally ice cold upon discharge of gas. This has a profound effect on the gas discharge behavior (slowing it). We will not be considering that aspect in this problem.[br][br][color=#000000]Suppose that a gas tank used as thrust leads to an acceleration function that looks like:[br] [center][math]a_x(t)=a_{max}e^{-\frac{t}{\tau}}[/math][/center][/color][/color]where [math]\tau[/math] is called a [b]time-constant[/b] for the decay, and [math]a_{max}[/math] is the initial (and maximum) value of the function. We wish to find the velocity as a function of time. Before doing that math, however, let's discuss the time-constant.[br][br]Time constants are used in lots of fields of study in which exponentials play into the math. In this case, the acceleration drops in an exponential way. It's worth wondering how quickly it drops. The first thing to notice is that with a bigger time constant, there is a more gradual drop in acceleration... or that it takes more time to decay. Sometimes without using numbers, it is meaningful to ask how much it is reduced in "one time constant" or "after three time constants", etc. Obviously the answer is that after one time constant we get [math]e^{-1}=\frac{1}{e^1}=\frac{1}{2.7181...} \approx 0.368 = 36.8\%.[/math][br]That means that after one time constant we have just over a third of the max acceleration. After two times constants we get [math]e^{-2}=\frac{1}{e^2}=\frac{1}{2.7181...} \approx 0.135 = 13.5\%.[/math] After three constant [math]0.0498 = 4.98\%[/math], after four [math]0.0183 = 1.83\%[/math], after five [math]0.00673 = 0.673\%[/math], after ten time constants there is basically nothing left. We're down to [math]0.0000454 = 0.00454\%[/math]. So while exponentially decaying functions never quite reach zero, they are really, really small after 5-10 time constants. In practice then, when I see a time constant of 10 seconds, I know that by 50 to 100 seconds, the process is basically over. [br][br]In the present context, it means that the acceleration due to my tank is essentially zero after 5-10 time constants. We'll see how that plays out after we work some numbers. So let's do the math. To find the velocity that's acquired from the thrust-induced acceleration given above, we need to use[br][br] [center][math]\Delta v_x = \int_0^t a_x(t)\;dt.[/math][/center][br][center][math]\Delta v_x=\int_0^t a_{max}e^{-\frac{t}{\tau}}\;dt.[/math][/center][br][center][math]\Delta v_x=-\tau a_{max}e^{-\frac{t}{\tau}}|_0^t.[/math][/center][br][center][math]\Delta v_x=-\tau a_{max}(e^{-\frac{t}{\tau}}-1).[/math][/center][br][center][math]\Delta v_x=\tau a_{max}(1-e^{-\frac{t}{\tau}}).[/math][/center][br][br]This result gives the acquired change in velocity due to the thrust. We need an initial velocity in order to have an expression for [math]v_x(t).[/math] Recall that [math]v_x(t)=v_x(t=0)+\Delta v_x.[/math] So if the object starts at rest from the frame of the observer, then they are the same. Let's look at a plot of the result below.

If you look at the plot above, you can adjust the time constant. With the default value of one second for the time constant, my eye says that terminal speed is reached around five seconds, or five time constants after the start. This is true regardless of the value of the time constant. So with a time constant ten times as large, you'd see the speed level off around 50 seconds. Since this sort of exponential behavior is seen all over nature, it's useful to remember this behavior.