ma006 ยป

Contents

Motion Problems with Currents

In the last lesson, we looked at Motion Problems, by looking at the distance, rate and times of different objects. We saw how fast some people could bike and we saw how fast some things could travel. We're going to tackle a similar Motion Problem, this time it's going to involve a current. And in order to solve these problems involving current, we want to be clear on 2 words, upstream and downstream. If we think about swimming laps inside of a pool, I could say that my speed would be 5 miles per hour. That's about how fast I swim from each side of the pool. 5 miles per hour would be considered to be my speed in still water, since there's no current pushing me forward or backward. I'm just swimming along inside of a pool. But what if instead I went swimming in a river? A river has a current to it, so if i swim downstream I would be swimming with the current. And if I swim upstream i would be going again the current. If the speed of the current is 3 miles per hour. And I know that in still water I can swim 5 miles per hour, what would be my speed if I was traveling downstream? And, what would be my speed if I traveled upstream? Write those 2 answers here.

Motion Problems with Currents

Well, if I'm traveling downstream, I can travel 8 miles per hour. And if I'm traveling upstream, I can only go 2 miles per hour. Good work if you figured these two out. One way to think about this is just by adding two vectors. We know that I can travel at 5 miles per hour. And a current's going to help me along. So, that's going to push me another 3 miles per hour. Now this only happens when I travel downstream, I get pushed along by the current. So, my total speed would be 8 miles per hour. But let's say I decide to swim upstream instead. Notice that the current is going against me. We're moving in opposite directions. This means that my rate of 5 miles per hour is slowed down by 3 miles per hour. So, I can only travel at 2 miles per hour.

Upstream and Downstream Rates

We'll use this knowledge of downstream and upstream to help us solve a motion problem with the current. We know that if we travel downstream we'll end up be travelling faster since we'll be going with the current. But if we travel upstream we'll have to be going slower since we're going against it. Let's see if we can solve this first motion problem together. You take a boat upstream for ten miles, and then back down for 15 miles. If the speed of the current in the river is five miles per hour, and you make the entire trip in one hour, let's see if you can find the speed of the boat in still water. Just like all the motion problems we've encountered up until this point, we want to use a table to help us organize all this information. So here's my table and we'll have the distance for the boat upstream, its rate and its time, and the distance of the boat downstream with that rate and that corresponding time as well. We want to find the speed of the boat in still water. So, let's make that be our variable. X is going to be the speed of our boat and in miles per hour. So now that we have a variable what expression can you write for the rate, or the speed of the boat, moving upstream, and the speed of the boat moving downstream.

Upstream and Downstream Rates

Moving upstream, the boat would go at x minus 5 miles per hour, and moving downstream, the boat would travel at x plus 5 miles per hour. Great thinking for finding these two. Now, this one was pretty tough, so don't worry if you didn't get it right. The key lies in the speed of the current. We know the speed is five miles per hour. So, if the boat travels at x miles per hour, then with the current, it can go five miles per hour faster, that would be moving downstream. But against the current, we know the boat travels slower, so it would lose the five miles per hour, since it's working against the current.

Motion Problem with a Current 1 Table

So, now that we have expressions for the boat's rate upstream and the boat's rate downstream. I want you to fill in this other missing information. Keep in mind that these 2 should be an expression. They should involve the distance and the rate.

Motion Problem with a Current 1 Table

You take the boat upstream for 10 miles, so we know this should be 10 and then when you travel back downstream, you go 15 miles. So, this box should be 15. Now to find the time of the boat, to travel 10 miles, upstream, we simply take the distance and divide it by it's rate. So, we'll have 10 divided by x minus divided by its rate. So we'll have 15 divided by x plus 5. Now we have all of our information and we want to try and set up an equation in order to try solve.

Motion Problem with a Current 1 Equation

Now that you have all the information about your boat moving upstream and downstream. I want you to write an equation that lets us solve for x. Take some time to think about it, and when you think you have it, type it in here.

Motion Problem with a Current 1 Equation

When you drove the boat upstream and downstream, it took a total time of one hour. This means that if we add the time to go upstream to the time to go downstream, those should add to one. So, we'll add the time to go upstream to the time to go downstream, and that should equal 1. Great thinking if you found this piece of information which you led you here.

Motion Problem with a Current 1 Solved

So, now that we have an equation for x, I want you to solve for it. What's the speed of the boat in still water?

Motion Problem with a Current 1 Solved

The boat travels at 25 miles per hour, great algebra solving if you found that answer. We know the lowest common denominator for this equation would be x minus 5 times x plus 5. So, we multiply our equation through by the LCD. These binomials cancel with one another, leaving us with this equation. Here on the right hand side I recognize that this was the difference of squares pattern. So I just have x squared minus 25 a squared minus b squared. Next we distribute to get this equation. Now we just want to combine my terms. 10 x plus 15 x equals or just remove the negative 25 since it's balanced on each side of the equation. This leaves us with 25 x Eequal to x squared. We have a quadratic equation, so we want to move this term to the other side, with the positive x squared term. This allows us to set our equation, equal to 0. Now we're in a position to factor this quadratic. I've rewritten it here, and then I factored an x out of each of these terms. So I'll have x times x minus 25. Now we set each factor equal to 0, leaving us with x equals 0, or x equals 25. So there are actually two possible values for x, 25 or 0. But since the boat is actually moving, we know it can't be zero. Therefore, the boat must move at 25 miles per hour in still water.

Motion Problem with a Current 2

Let's see if we can apply our same strategies to this motion problem. In 4 hours Katrina can go 15 miles upriver and come back. The speed of the current of the river is five miles per hour. What is Katrina's speed? Just like from before, we're going to set up a table to list all the information we know, and write expressions for anything that might be missing.

Motion Problem with a Current 2

So, we're looking for Katrina's speed alone, that would be without the current. We know when she goes upriver or upstream, she goes 15 miles. So this distance is 15. She has to come back to her original starting point, so that means she has to go another 15 miles. To get back where she started. That would be her distance downstream. We know the speed of the current is 5 miles per hour, but we don't know Katrina's speed. It's x. So, if Katrina is travelling downstream, she travels faster. So that would be x which is our speed plus the current which is 5. If Katrina is traveling upstream she's fighting the current, so she's going slower. So that would be x minus 5. Whenever we work with these motion problems and a current. We'll always add or subtract the current to whatever speed our person or object is going. So in this case Katrina's speed was x. So we need to add the current for her downstream speed and subtract the current for her upstream speed. Now we're ready to find expressions for the time. We know the time is distance divided by rate. So going upstream we'll have 15 divided by x minus 5. And going down stream we'll have 15 divided by the quantity x plus 5. Now this was pretty tough, so if you got at least part of this right, great work.

Motion Problem with a Current 2 Equation

Notice that we weren't given the actual time for Katrina to travel upstream or the actual time for her to travel downstream. We just have expressions for each of these. But we do know something about the sum or the total amount of time she spends on the river. So I want you to think about what equation you can write in order to solve for x. When you think you've got it enter it here.

Motion Problem with a Current 2 Equation

This is the equation we can write. Again, the key here is the total amount it takes for Katrina to go upstream and downstream is 4 hours. So, if we add her time upstream to her time downstream, those should sum to 4.

Motion Problem with a Current 2 Solved

So now that we've set up our information and found an equation, we're ready to solve for x. What do you think is Katrina's speed?

Motion Problem with a Current 2 Solved

Katrina travels at 10 miles per hour. Great work if you found this number. We can start solving this equation by multiplying through by the LCD. The lowest common denominator in this case is x minus 5 times x plus 5, the product of these two denominators. When we multiply we'll get 15 times x plus 5 here plus multiplied these two binomials together since we have a minus b times a plus b. That's a squared minus b squared, our difference of squares pattern. Next we distribute the 15 here and here and we distribute the positive 4 to get this new equation. Combining like terms on the left side of the equation, we'll have Next we want to try and factor this quadratic equation. So we'll need to move this 30x over to the right hand side in order to set the quadratic equal to 0. I'll list the 4x square term first then the negative 30x term and then negative Here we'll want to use factoring by grouping in order to factor this quadratic. We find factors of 2 times negative 50, which is negative 100 that sum to negative 15. Negative 20 and positive 5 multiply to give us negative 100 and they sum to negative 15. So we'll use these two numbers to rewrite negative 15x as negative 20x plus 5x. Then we just use factoring by grouping to get two factors of 2x plus 5 and x minus 10. We set each of these factors equal to 0 so x can equal 10, or x could equal negative 5 halves. We know x really represents Katrina's speed. So, we can't use the negative value. It's invalid. This is how we know Katrina travels at 10 miles per hour.

Motion Problem with a Current 3

Here's our third problem with motion. This time we have a speed boat that takes trip for the boat would be upstream and the boat's going to travel at its same speed. But against the current and that's going to take it 8 hours. Your challenge is to find the speed of the boat in still water. As always, we'll use a table to help us get started. So again here, we're looking for the speed of the boat. So we'll let that be x. See if you can fill in the missing information for our table. And this time there is something different with our time column. We're actually given the times to go upstream and downstream. This means that we'll have numbers here and expressions for distance here. See if you can figure this one out.

Motion Problem with a Current 3

It takes the boat 6 hours to go downstream so we know what this corresponding time must be 6. We know the speed of the boat going downstream is faster with a current so we'll have the speed of the boat, which is x. And then we'll add the current which is 12. The return trip for the boat is going to be 8 hours so for upstream or going against the current we'll have eight for the time. The speed of the boat against the current will be x minus 12. We'll have the speed of the boat which is x and then we'll have to subtract off the current speed which is going upstream or downstream. We'll need expressions for each of these. The distance is equal to rate times time. So when going upstream the boat's distance would be x minus 12, the rate times 8, the time. And when going downstream we'll have the rate x plus 12 times the time which is 6. This is our complete table with all the information we need. And if we want we go ahead and simplify each of these expressions. When we distribute the eight, we'll have 8x minus 96 for the first one. And we distribute the six, we'll have 6x plus 72 here.

Motion Problem with a Current 3 Equation

Now that we have this information, I want you to think carefully about what equation you can write in order to solve for x. Think about what is true for this problem. Think about the distances, the rates, and the times. You'll be able to relate one of these things together based on the information in the problem.

Motion Problem with a Current 3 Equation

The key here is that our boat is going to make a return trip. The boat's going to start and travel downstream and then turn around and come right back upstream. We know that the distance that the boat travels downstream is the same as the distance for the boat to travel back upstream. This means that these expressions are these two numbers, must be equal. So we'll know 8x-96 equals 6x plus 72. This is our equation that we can use to solve.

Motion Problem with a Current 3 Solved

So now what I want you to do is use this equation and solve for x, the speed of the boat. Write your answer here.

Motion Problem with a Current 3 Solved

In order to solve for x, we'll subtract 6 x from both sides. Then we'll add 96 to both sides, and finally we'll divide by 2 to get x alone. So one x is equal to 84. Great solving if you found 84 miles per hour for the speed of the boat.

Motion Problem with a Current 3 Distance

Keep in mind that there are other questions that might be asked. For example, now that we know x, we can go back and find other values. We could find the speed of the boat against the current, the speed of the boat with the current, or the distance travelled. For this question, I just want you to tell me, what's the distance travelled in each direction. Type that answer in here.

Motion Problem with a Current 3 Distance

The boat travels 576 miles in either direction. Great work if you found this. We can find this answer by plugging in the value of x which is 84 and take each of these expressions. Either expression will give us the correct answer, so let's make sure. If we take the expression for distance for upstream and replace the value of x with 84, we see that the answer is 576. The same would be true for downstream. We replace x with 84 to get 504 plus 72 which equals downstream first and then it went back the same distance to return. So, this supports our reasoning for how we solve the equation and we know the boat traveled 576 miles in each direction.

Motion Problem with a Current 4

Let's try working through 1 last problem with motion and a current. In this problem, a riverboat averages 12 miles per hours in still water. It takes the boat an hour and 4 minutes to go 6 miles upstream, and then come back. What I want you to do, is to find the speed of the current. Before we find the speed of the current, let's set up our information. I want you to fill in each of these parts of our table, and be sure to use the variable x for your expressions. Think carefully about which of these should be numbers, and which of these should be expressions.

Motion Problem with a Current 4

We know the boat travels upstream for 6 miles so here this is 6. The boat is going to travel the same distance to return back from where it started, so we know the downstream distance must also equal 6. We know the boat travels 12 miles per hour in still water. So that means with the current, we would add x going downstream and that means going against the current. We would subtract x, that would be moving upstream. The key idea is that for the boat's rate, we let the speed of the current equal x. So the boat would be traveling faster going downstream, and slower going upstream. And finally, for our time column, we know that we just need to take the distance, and divide it by the rate. So, going upstream, the boat would take 6 divided by 12 minus x hours. And going downstream, the boat would take 6 divided by 12 plus x hours. Fantastic work if you've done all these expressions, this is pretty tricky.

Motion Problem with a Current 4 Time

If we add these two times together we know they should sum to the total amount of time it takes our boat to travel 12 miles. The boat can make the trip in 1 hour and 4 minutes. But here's our problem, this is in hours and minutes and these units are in only hours so we can't quite set up an equation just yet. We need to convert this to only units of hours. So if we want to convert this to hours, what improper fraction could we write?

Motion Problem with a Current 4 Time

Well, we know 1 hour and 4 minutes would be hour and 4 60ths of an hour. 4 minutes is just 4 60ths of an hour, since an hour can be broken up into 60 minutes. 4 60ths simplifies to 1 15th, since there's a common factor of four and the numerator and in the denominator. And then finally to add these two numbers together, we can change one to be 15 15th. This gives us a total sum of

Motion Problem with a Current 4 Equation

So now we know one hour and four minutes really equals 16 15ths hours. So, now our total time is in only the units of hours. Knowing this, what equation can we write to solve for x? The equation needs to involve the time going upstream, the time going downstream, and the total amount of time.

Motion Problem with a Current 4 Equation

this is the equation that we can write. If we add the amount of time to go upstream, the amount of time to go downstream. That should sum to our total time which is 16 15th hours. Nice work if you figured that out.

Motion Problem with a Current 4 Solved

So now that we have an equation. I want you to solve for the speed of the current. Solve for x.

Motion Problem with a Current 4 Solved

The current travels at 3 miles per hour. Excellent problem solving if you found that answer. To solve this equation, we'll start by multiplying through by the lowest common denominator. The LCD in this case is 12 minus x times 12 plus x times 15. The product of all three of the denominators. So we'll multiply each of these terms by the LCD. We'll note the factors of 12 minus x simplified in the first fraction. 12 plus x simplifies in the second fraction. And 15 simplifies in our last fraction. Each of these reduces to 1. This leaves us with this equation. We know 6 times 15 equals 90, so we'll change these. And then we know this is a difference of square pattern. a minus b times a plus b. So this will equal a squared minus b squared. 12 squared is 144. We'll have our minus sign in between the terms, and x squared is x squared. Next we'll distribute 90 to each of these terms. We'll distribute 90 here, and we'll distribute positive 16 here. We get this equation, and then we want to combine the like terms on the left. We know positive 90x, and negative 90x sum to 0. And 1080 plus 1080 sums to 2160. And the right hand side of our equation remains the same. Now that we have this equation, we want to try and solve for x. Since I only have an x squared term and no other x terms, I want to move this constant to the left hand side, and then divide through by negative 16. So we can take the square root of both sides. We subtract 2,304 from both sides to get negative 144 equals negative 16 x squared. Then we divide both sides by negative 16 to get 9 is equal to x squared. We take the square root of both sides so x can equal positive 3 or negative 3. Since x is the speed of the current, we know that we need the positive value. x must equal 3 miles per hour.