# Video: Finding the Limiting Speed of an Object Subject to a Resistive Force Proportional to the Speed of the Object

A boater and motor boat are at rest on a lake. Together, they have mass 2.0 × 10² kg. The thrust of the motor provides a constant force on the boat of 40 N in the direction of the boat’s motion. If the value of resistive force of the water in newtons is two times the value of the speed of the boat in m/s, find the limiting speed of the boat.

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### Video Transcript

A boater and motor boat are at rest on a lake. Together, they have mass 2.0 times 10 to the two kilograms. The thrust of the motor provides a constant force on the boat of 40 newtons in the direction of the boat’s motion. If the value of resistive force of the water in newtons is two times the value of the speed of the boat in meters per second, find the limiting speed of the boat.

Let’s highlight some of the critical information we’re given in this problem statement. First, we’re told that the combined mass of the boater and the boat is 2.0 times 10 to the two kilograms. And we’re also told that the motor of the boat exerts a constant force of 40 newtons in the direction of the boat’s travel. And lastly, we’re given this very interesting relationship of the resistive force of the water on the boat. For shorthand, let’s call that force 𝐹 sub 𝐷 for drag force. And we’re told that this drag force is equal to two times the value of the speed of the boat in meters per second. In other words, we could write the drag force as two times 𝑣, where 𝑣 is the speed of the boat in meters per second, and this force is in units of newtons. So with this drag force and with this information about our situation, we’re asked: What is the limiting speed or the maximum speed of the boat?

Let’s draw a picture of our boat on the water in order to start working towards this solution. Here, we have our boat and our boater on the water and we’re told that there is a force that pushes the boat forward, the force from the motor that we’ve called 𝐹 sub 𝑚, and that that force always points in the direction that the boat is moving. We’re also told that in the horizontal direction, there is a resistive force of the water that acts backward on the boat, and we’ve called that 𝐹 sub 𝐷, the drag force. And we’ve been given an expression for 𝐹 sub 𝐷 that relates it to the speed of the boat, that 𝐹 sub 𝐷 is equal to two times the boat speed in newtons. Now you’ll see that there are vertical forces that act on the boat as well. There is the weight force of the boat acting down, and there’s the force of the water on the boat pushing it back up, a normal force. But for the purposes of our question, we can neglect or disregard those vertically oriented forces and just focus in the horizontal direction. Now we’ve been asked for the limiting speed of the boat. In other words, what is the maximum speed that the boat can achieve under these conditions. We’ll call that speed 𝑣 sub max.

Now let’s picture what’s going on here in our minds a little bit to get familiar with how this speed and drag force relationship works. We’ll make a table that shows what the speed of the boat is and what the resulting drag force is. Now once again, we have this given relationship between the speed of the boat, 𝑣, and the drag force on the boat, 𝐹 sub 𝐷. Now we’re told that the boat starts from rest. So when 𝑣 is zero meters per second, then at that time, the drag force on the boat is zero, two times zero is zero. Now say the motor of the boat engages, and the boat starts to speed up. Now a few moments later, say that the speed of the boat is two meters per second, then the drag force on the boat is no longer zero; it’s increased. And now it’s two times two or four newtons. Now as the boat continues to gather speed, say that it’s speed becomes five meters per second, well then at that moment, by our drag force relationship, the drag force on the boat is two times five or 10 newtons.

So you can see as this goes on, the speed of the boat continues to increase, and the drag force increases at the rate of two times the boat speed. So it would seem from this table that the speed of the boat has no upper limit, but indeed it does. And that has to do with 𝐹 sub 𝑚, the force of the motor. We’re told that that force is a constant value of 40 newtons. So the boat won’t continue to speed up indefinitely. Eventually, the thrust that pushes it forward in the water will be balanced out by the resistance of the water on the boat. In other words, eventually when the boat is at its maximum speed, when the speed equals 𝑣 sub max, then in that case the magnitude of the force of the motor will be equal to the magnitude of the drag force. This is the condition that will let us solve for 𝑣 max, the condition when these two forces balance one another out, and the boat is no longer accelerating.

Now all this talk about forces and acceleration and 𝑣 max may remind you of one of Newton’s laws of motion. We’ll take advantage of that law to solve for the maximum speed the boat can attain, and that law is Newton’s second law. And the second law states that the net force acting on an object, in our case a boat, is equal to the mass of that object multiplied by its acceleration. So for us, the net force in the horizontal direction on our boat is equal to the mass of the boat and the boater multiplied by its acceleration.

But before we go further, recall this condition that we’ve identified for 𝑣 sub max. We’ve said that when the speed of the boat is at its maximum, the two forces in the horizontal direction balance one another out. So let’s write out a revised version of the second law for our situation, under the condition that the speed of the boat is at its maximum or limiting speed. First, let’s choose a direction to be positive in our diagram. We’ll say that motion to the right is motion in the positive direction, and motion to the left is therefore negative. So with that convention, we’ll say that the two forces in this direction, 𝐹 sub 𝑚 and 𝐹 sub 𝐷 arranged in Newton’s second law, will look like this: 𝐹 sub 𝑚, the motor force, minus 𝐹 sub 𝐷, the drag force, when the speed of the boat is at its maximum, is equal to zero. The reason that this equation is equal to zero is because our acceleration is zero when the velocity is at its maximum. It’s no longer increasing, but it’s rather steady. And therefore, 𝑎 is zero, so 𝑚 times 𝑎 is zero.

Now let’s rewrite this equation in terms of 𝐹 sub 𝑚 and 𝐹 sub 𝐷 in as much detail as we can. We’re told that 𝐹 sub 𝑚 is a constant of 40 newtons, and we’re also given that 𝐹 sub 𝐷 is equal to two times the speed of the boat or two 𝑣. So 40 newtons minus two times 𝑣 is equal to zero. Now recall that for this equation to be true, the 𝑣 that we’re talking about, the speed of the boat we’re referring to, is the maximum speed the boat can attain. So actually, the 𝑣 in this equation is 𝑣 max.

Now let’s rearrange this equation to solve for 𝑣 sub max, the highest speed the boat can reach. First, we can add two times 𝑣 sub max to both sides of the equation. That term then cancels out on the left side of our equation. And if we then divide both sides of our resulting equation by two, the factor of two cancels out on the right side. And we’re left with an equation that states: 𝑣 sub max is equal to 40 newtons divided by two.

So we see that the maximum speed the boat will attain is 40 divided by two meters per second or 20 meters per second. This is the limiting speed or the maximum speed the boat can reach under these conditions.