(III) A motorboat traveling at a speed of 2.4 m/s shuts off it...

Question

(III) A motorboat traveling at a speed of 2.4 m/s shuts off its engines at t = 0. How far does it travel before coming to rest if it is noted that after 3.0 s its speed has dropped to half its original value? Assume that the drag force of the water is proportional to v.

Accepted Answer

Hello, let's go through this practice problem. A speed about moving at 4.2 m per second, turns off its engines at T equals zero. If its speed decreases by half, after two seconds, calculate the distance it covers before stopping. Presume the water's drag force is directly proportional to V option A 10 m B 12 m C 17 m or D 22 m. All right. So let's begin by drawing a diagram just to get a feel for what's happening in the problem. So we have a boat moving in the water. So I'll represent the boat with this square like shape and it's floating through the water and it's moving at some speed. We say V now the problem tells us that the speed boat turns its en engines off. So at T equals zero, the only force acting on the boat is the drag from the water. And remember that drag opposes the motion. And we're told that that drag has proportionality with V. So we'll say that this drag force has a magnitude of BV where B represents the speed and B is some constant. So we can set up a pretty simple force equation using Newton's second law or the only force acting on the boat is negative BV. And according to Newton's second law, the sum of all the forces acting on an object are equal to the mass of the object multiplied by its acceleration. Now, in order to find a relationship between the speed and the time we're going to want to set up a differential equation. And we can do that using the fact that acceleration is defined as the rate of change or the derivative of velocity with respect to time. So this force equation can alternatively be written as M multiplied by the derivative of speed with respect to time is equal to negative BV. So we now have what is known as a separable differential equation where in order to find a relationship between speed and time, we need to first multiply both sides of the equation by DT so that we can get the two differential variables on opposite sides of the equation. So we're gonna bring the DT over to the right side of the equation and then divide both sides of the equation by V to bring the V over to the left side of the equation where the DV is. So we're gonna say DV divided by V is equal to and we're gonna bring the M over to the right side of the equation as well to make things even easier. So negative B divided by M all multiplied by DT, so now we have the two differential variables on either side of the equation. And in order to get rid of those differential variables and just have an equation connecting V and T, we're going to have to integrate both sides of the equation. So we're going to add integrals. The bound of integration for time is just going to be from zero to whatever the final time is, we're just going to say is T and the bound of integration for speed is just gonna be from the initial speed V not to the final speed which we'll just call V. Fortunately for us, both of these integrals are fairly easy to do the right integral with the time is very simple because there is no T in the thing we're integrating, we're just integrating one basically. And then we multiply the negative B divided by M. So we just materialize A T basically. So negative B T divided by M. And because the lower bound of integration is zero, you don't even need to worry about the fundamental theorem of calculus. The left hand integral with the DV divided by V is a little more involved, but it's still barely easy. The integral of one divided by the variable of integration is just equal to the natural log of the absolute value of the variable. And when we apply the fundamental theorem of calculus over a natural log function, all we really have to do is just divide the upper bound by the lower bound within the natural log. So the integral is equal to the natural log of V divided by V. Not. So our final little equation here is the natural log of V divided by V not equals negative BT divided by M. To actually get a relationship between V and T. To find a formula for speed with respect to time. We're gonna get, we're gonna wanna get those vs outside the natural log. So we're gonna raise both sides of the equation uh as an exponent of E. So left hand side of the equation becomes V divided by V not. And the right hand side of the equation is E raise to the power of BT divided by M. And then we multiply both sides of the equation by V not so V is equal to V not multiplied by E raise the power of BT divided by M. Now, according to the parameters set by the problem, we're specifically looking at the case where the final speed is half of the initial speed. So where V is equal to V not divided by two. So in this new version of the equation, the V knots will cancel out. So we're just left with one half is equal to E raise the power of negative B multiplied by the time. And again, the problem specifies that this half speed thing happens and where when T is equal to two seconds fight by M. Now remember that the problem is actually asking for a distance. So we're going to need to set up another integral because we can find position by integrating velocity. So if we look at our speed equation notice that it is a formula for speed, which is the derivative of position with respect to time, so we can set up a differential equation related to this or DX divided by DT is equal to V. Not E raise the power of negative BT divided by M were going to integrate both sides of the equation after setting it up as a separable differential equation. So we're gonna multiply both sides of the equation by DT. So we just get DX on one side equals V not E raise the power of negative BT divided by M multiplied by DT. Now we can integrate both sides of the equation where the bounds of integration for position are from zero, the starting position to X and the bounds of integration for time are again just gonna be zero to T. The left hand side integral is fairly simple. It's just X because we're integrating nothing from zero to X. And on the right hand side of the equation, we apply a basic chain rule integration and we find that side to be equal to V not multiplied by negative M divided by B all multiplied by E raised the power of negative BT divided by M minus one in parentheses. Distributing out the minus signs. A better way to write, this would probably be V not multiplied by M divided by B multiplied by one minus E race. The power of negative BT divided by M. Now, the problem is asking us to find how far the boat ultimately travels as it is slowly decelerating, slowly slowing down before it stops. So we're essentially looking for the place where at the very end of its motion to where time the time variable goes to infinity. So notice of the term containing the time variable is an exponential where the exponent itself is negative when the exponent is negative. That is the same thing as if we have one divided by the exponential with a positive exponent. So if T is going to infinity, and that means the denominator of this far right term is going to zero. So at that limit, this entire term disappears, which means we're just left with one minus zero, which is just equal to one. And the whole equation just simplifies down to this much more nice to look at and much smaller term of V not M divided by B. So that's fine. And the problem statement does tell us what the initial speed is, but we still need to figure out what M divided by B is. And that's what we're going to consider the condition we discussed earlier that where the final speed is one half of the initial speed. Because if you'll notice from the equation we wrote down up here, we can actually rewrite this equation algebraically to solve it or be divided by M. First, we need to get this exponential out of the way. So we're gonna take the natural log of both sides of this equation. So the natural log of one half which can be more simply written as its inverse. So the negative of the natural log of two is equal to negative B multiplied by two seconds divided by M. These negative signs can go away. So we have the natural log of two equals B multiplied by two seconds divided by M. And then we'll divide both sides of the equation by two seconds. So we just have the B to M ratio is equal to the natural log of two divided by two seconds. So now we're going to plug this into our X formula. So X is equal to V not, which is given in the problem as 4.2 m per second multiplied by the M to B ratio, which is equal to the opposite of the B to M ratio we just found. So instead of writing the natural log of two divided by two seconds, instead, it's going to be two seconds divided by the natural log of two. And if you put this into a calculator, then we find a position of approximately 12 m and so 12 m is our answer, which corresponds to option B so option B is the final answer to this problem and that is it for this problem. I hope this video helped you out and please consider checking out our other tutoring videos which will give you more experience with these types of problems. That's all for now and I hope you have a lovely day. Bye bye.

(III) A motorboat traveling at a speed of 2.4 m/s shuts off its engines at t = 0. How far does it travel before coming to rest if it is noted that after 3.0 s its speed has dropped to half its original value? Assume that the drag force of the water is proportional to v.

Key Concepts

Kinematics

Drag Force

Exponential Decay

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