A skydiver jumped out of a stationary helicopter. His terminal velocity was 55 meters per second. Given that the air resistance is directly proportional to velocity, how long did it take for his speed to reach 54 meters per second? Round your answer to one decimal place. Take 𝑔 which is the acceleration due to gravity to equal 9.8 meters per second squared.
This question is all to do with motion. To begin, we’re going to draw a very simple diagram. The question describes a skydiver which we’ll represent using this circle. Now, we know that the weight of an object points directly downwards towards the Earth. As soon as our skydiver jumped, he will begin to fall in this direction since this is the direction of the force of his weight. Here, we have represented the weight of the skydiver as 𝑊. And we know that this is equal to the mass of an object 𝑚 times the acceleration due to gravity 𝑔.
Now the system given in our question involves another force and that is air resistance. Air resistance will oppose the direction of motion of an object. In our system, it’s important to know that our skydiver has jumped out of a stationary helicopter. This means he begins with a velocity of zero. Now, we’ve already said that the weight of the skydiver is acting downwards, which means he’ll begin to accelerate in this direction. In other words, his velocity will begin to increase in this direction.
We have also said that air resistance opposes the direction of motion. What this means is that air resistance, denoted as 𝑅 here, will be a force acting vertically upwards. Since these are the only two forces acting on our skydiver, we can safely assume that the entire phase of motion will occur in the vertical direction. And our skydiver will never have any horizontal velocity. Of course, in the real world, there would be many other factors, such as wind, which would mean this is not the case. However, here we’re dealing with an idealized system.
Okay, the next piece of important information is that the air resistance is directly proportional to velocity. We could write this as 𝑅 is proportional to 𝑣, the velocity. Another way to express this, however, is that 𝑅 would be equal to some constant 𝑘 times 𝑣. This would be more useful in the equations that we’ll be using.
Let’s begin to form these equations now. Our skydiver will have a net force acting on him. We know that the net force felt by any object is equal to its mass times its current acceleration. We also know that in our system, the net force is the result of the weight and of air resistance. If we define the downwards direction as the positive direction, we can say that the net force is equal to 𝑊 minus 𝑅. Let’s express these in terms of the two equations we already formed. 𝑊 is equal to 𝑚𝑔 and 𝑅 is equal to 𝑘𝑣. This allows us to form another more useful equation. 𝑚𝑎 is equal to 𝑚𝑔 minus 𝑘𝑣.
The reason this equation is useful is because the two variables involved are acceleration and velocity. Both of these variables involve motion. All of the remaining terms in this equation are constants. We have 𝑚, the mass of the skydiver, 𝑔, acceleration due to gravity, and 𝑘, the constant we define for air resistance.
For our next step, we’ll divide both sides of the equation by 𝑚, allowing us to isolate acceleration on the left-hand side. This gives us that 𝑎 is equal to 𝑔 minus 𝑘 over 𝑚 times 𝑣. We’ll see why this is useful in a moment. This equation allows us to incorporate the next useful piece of information given by the question. And that is the terminal velocity of the skydiver is 55 meters per second. Terminal velocity is the maximum velocity that the skydiver will reach during the phase of motion. That means his velocity will increase up to this value but then it will not increase anymore. It will remain constant. We know that when the velocity of an object is constant, that must mean it’s not accelerating. So its acceleration is zero. This allows us to substitute some values into our equation.
At terminal velocity, which we’ll call 𝑣 max, the acceleration will be zero. And the question gives us that 𝑣 max is equal to 55. Our equation then becomes zero is equal to 𝑔 minus 𝑘 over 𝑚 times 55. At this stage, for clarity, we can also substitute in the known value for 𝑔 into our equations, which is 9.8. Finally, let’s rearrange this equation. We’ll do this so that we have the unknown constants on the left-hand side and our numerical values on the right-hand side. Hopefully, we can see that we have just found a numerical substitution for this 𝑘 over 𝑚 term in our original equation. Substituting this in gives us that acceleration is equal to 9.8 minus 9.8 over 55 times 𝑣. Here, we’ll put the useful values that we used to one side and move on.
We have now reached the crux of this problem. Here, we should remember that in this system, and indeed many complex systems, acceleration and velocity can be thought of as time-dependent variables. Going one step further, we should remember that acceleration can be thought of as the rate of change of velocity with respect to time. Using the language of differential equations, in shorthand, this can be written as d𝑣 by d𝑡.
Okay, we now might realize that what we have formed here is a differential equation. We have a normal velocity term on the right-hand side. But on the left-hand side, we have the first derivative of velocity with respect to time. Let’s clear some space to make room for further calculations. What we have formed is a first-order separable ordinary differential equation. Now, we won’t go too far into the definition of this. But this separable part is important. In our case, what this means is that our derivative d𝑣 by d𝑡 can be expressed in terms of some function of 𝑣 multiplied by some function of 𝑡.
For our equation, we see that no 𝑡 terms exist, so this condition must obviously be satisfied. Again, the full details of this are slightly outside the scope of this video, but it does allow us to use the following method. First, we’ll divide both sides of our equation by 9.8 minus 9.8 over 55 𝑣. In essence, what we’re doing here is moving our function in terms of 𝑣 over to the left-hand side of our equation. For the next step of this equation, you might often see methods which treat d𝑣 by d𝑡 as a fraction. The equation is then rewritten like so and we would ambiguously integrate both sides. Let’s undo these steps and follow a slightly more rigorous route.
For our next step, we integrate both sides of our equation with respect to 𝑡. Next, we recall that the chain rule allows us to say that d𝑣 by d𝑡 d𝑡 is simply equivalent to d𝑣. Once we perform a substitution, we reach the same step as the previous shorthand. Okay, let’s now evaluate these integrals. As it turns out, the integral on the left-hand side will end up a little less messy if we force the coefficient on our 𝑣-term to be one. We do this by multiplying the top and bottom half of our quotient by 55 over 9.8. We can then take this constant value outside the front of our integral and simplify the bottom of the quotient. While we’re in the process of simplifying, as a side note, we don’t really need this one on the right-hand side of our equation. It is far more common to simply write the integral of d𝑡.
Okay, let’s clear some space keeping our original differential equations since this was an important step. Let’s now move forward by solving these integrals. You may be able to solve the integral on the left directly, but for the sake of completeness, we’ll perform a quicker 𝑢-substitution. This will be 𝑢 equals 55 minus 𝑣. We can move forward by saying that d𝑢 is equal to negative d𝑣, or equivalently negative d𝑢 is equal to d𝑣. When performing these substitutions, we should be careful not to forget about this negative term which has been placed in front of our integral along with the constant terms. After doing so, we are left with the integral of one over 𝑢 with respect to 𝑢. And this is a standard result.
Of course, the result is the natural logarithm of 𝑢. We’ll now substitute 55 minus 𝑣 back in for 𝑢. Note that the right-hand side of our equation is a trivial calculation. The integral of d𝑡 simply gives us 𝑡. You might also notice that on both sides of our equation, we have not forgotten about the constant of integration 𝐶. Since these are arbitrary constants, we can easily combine them by saying that another constant, 𝐶 three, is equal to 𝐶 one minus 𝐶 two. Subtracting 𝐶 two from both sides gives us the following result. Negative 55 over 9.8 times the natural logarithm of 55 minus velocity plus a constant is equal to time.
At this stage, we have reached the general solution to our differential equation. However, we need to find the specific solution for our particular case. And we’ll do so by defining the value of 𝐶 three. And this can be achieved by considering an implied boundary condition or initial condition given by the question. Remember that velocity is a time-dependent variable. However, our question tells us that the skydiver jumped out of a stationary helicopter. This means that at time zero, the skydiver is also stationary. In other words, his velocity equals zero.
This allows us to substitute 𝑡 equals zero and 𝑣 equals zero into our equation. Great, we can now rearrange this. And we find that our constant 𝐶 three is equal to 55 over 9.8 times the natural logarithm of 55. Putting this together by substituting this value back into our original equation give us the specific solution to our differential equation. We’re gonna switch things around and write our equation in terms of 𝑡 since this will eventually be what we need to find. This is great. We now have an equation for time in terms of only velocity. We can tidy things up a little bit first using the common factor and then using laws of logs.
Now for the final stage, the question has asked how long it takes for our skydiver to reach a speed of 54 meters per second. Given that our skydiver is only moving in one dimension and his direction does not change, we can simply consider this 54 to be the magnitude of his velocity. The question is then, what is 𝑡 when 𝑣 equals 54? We can find this using our equation for 𝑡 in terms of velocity. We substitute in our value of 𝑣 equals 54, we then simplify, and we reach our answer.
Finally, we know that our question wants our answer rounded to one decimal place. And since the velocity was given in terms of meters per second, our time will be in terms of seconds. Our answer is then that it takes the skydiver approximately 22.5 seconds for his speed to reach 54 meters per second.