https://youtu.be/4U_-H6LT8Qc
PHYS 1101: Lecture Five, Part Six
Here we have a sports car picking up speed. It passes between two markers in a time of 4.1 seconds. The markers are separated by 120 meters. All the while, the car is accelerating by 1.8 meters per second squared. What is its speed at the second marker?
Here are steps. Visualize, what object? It’s going to be the sports car. Start to end. The start to end would be this passing between the two markers. Marker 1 is going to be the start and marker 2 would be the end.
What does the motion diagram look like? Let me sketch that above here. First here is my car, and that’s such a well-drawn car. I’m going to copy it over here at my finish. I’m sure you’d agree. The motion diagram now. Is this car starting from rest at the first marker and speeding up? Can I make that assumption? I don’t think I can. It says that the car is picking up speed as it passes between the two markers. I think it’s already going at some initial velocity and it’s then the velocity is definitely increasing, but that it has some initial velocity. I think I have to draw this for the motion diagram. My acceleration vector, just for reference, it’s acceleration, it’s increasing speed, same direction.
That’s a good initial picture of the scope of my problem. Now let’s be more specific and follow step two and find an axis and an origin and start thinking about our variables. The convention we picked for the previous problem would work well again here. So here I’m going to put my origin at the first marker. Here’s my second marker. Here’s my object we sketched at these two instances.
Okay, draw an initial velocity vector and draw the final just so you solidify in your mind whether they’re 0 or not, if they’re getting smaller or bigger. I will go ahead and draw an initial and a final. Because I’m near the edge of the page here I’m going to move that over a little bit, and I’m going to make my next and last velocity vector clearly bigger.
Think about these variables: time, position and velocity at each of these instances. Start my clock at 0 as usual. My initial origin. My origin is where the object is at that initial instance, so it’s positioned at 0. Do I know its initial velocity? I don’t. It’s positive, but I don’t know a number for it. Time, position and velocity at the end.
Time to cross the second marker, it looks like I do know that. It took 4.1 seconds. What’s its position coordinate at this instant given this is our origin? It looks to me like it’s going to be at the x-coordinate of +120 meters.
Then do we know our final velocity, the velocity at this instant? No we don’t. In fact, that’s what we need to know. This is what we want. If we get this final velocity, we know that the magnitude of that, just the positive value will be the speed and that’s what we have to type in, just a positive number meters per second for our answer.
The last variable to consider is a. Do we know what the constant acceleration is between the start and the finish, the two markers? You know what, we do. It’s 1.8 meters per second squared.
Okay. The next step here is to make our list of knowns and the variable we solved for. In fact, it would be a little bit redundant to repeat that entirely. Well, I don’t want to copy it. I’m just going to move it down a little bit and note that this really is what we’ve done by jotting down these quantities as we thought through setting up our problem. I am going to focus, though, or help me to focus by clearly saying that the variable I want is v, the final velocity.
Next step, pick equations and algebra our way to a solution. Let me go grab again a copy of our equations, our starting equation. All right, here they are. The first thing I’m going to do is go in and highlight the variable I want, v. Here it is, and here it is.
Next thing, I’m going to highlight the variables that we know and that we have. Do we know v0? No. Do we know a? Yes. And we know t. We know a and t. We know the position coordinates at the beginning and the end. It looks like that’s it. The variables that are not highlighted I don’t know and I need to solve for v.
My eye is really drawn to using this one, but I don’t know the initial velocity. You know what, I need that initial velocity for any of these equations. Here’s a strategy. What do you think of this idea? I don’t know the initial velocity but you know what? If I were to solve for the initial velocity and substitute it in, say, to this equation or this one, I would then remove that initial velocity from my equation. Let me just show you what I mean.
Let’s take equation one. Let me just write here, strategy. This is where there is no necessarily right or wrong answer. There are different ways to solve this problem. You just have to be sure whatever direction you head, you always mathematically do legitimate steps and you may find you head in a direction that doesn’t get you anywhere and you need to go back to the beginning and try a different strategy.
I’m just first going to start out thinking about equation one and realize that I don’t have any information about v0. For me to use any of the other equations, I really need to get rid of this variable v0. So let’s solve for v0 and then use this to get rid of v0 in another equation.
How do I? Let’s solve this for v0. In order to do that I’m going to subtract at from both sides. I’m going to end up with v minus at is equal to v0 plus at minus at, which is how I’ve moved it over to the left hand side. So if I just rewrite that as I would read it, v0 then equals v minus at. By solving for v0 now, I know I can replace v0 anywhere with its equivalent, what shows up here on the right hand side. These are all variables for which I have numbers for, so by substituting this into, let’s say, this equation, I will have gotten rid of this variable I don’t know.
I’m going to put here sub into equation two. You could have as easily substituted into equation three if you wanted. My eye is drawn to two. Let’s first start with equation two and I want to remove any terms that are 0 just so my algebra . . . it looks a little simpler in my equation that I’m working with. Let’s see, my initial marker and therefore my initial coordinate was 0. Looks like that’s the only one I can get rid of. x is equal to v0t plus one-half at2.
Now I’m going to do my substitution. We can move our next problem down. So for v0 I’m going to substitute in v minus at. x then is equal to v minus at, that’s for this. I still have to multiply by t, and then I have to add one-half at2.
I’m after the variable v, meaning I need to rearrange this so I have v is equal to blah, blah, blah, variables for which I have values for. So I need to do some algebra here. In order to isolate v I first have to multiply in this t. vt minus at2 plus one-half at2. I have to, in order to isolate it, I need to get v related to the other variables in terms of a sum or a difference so I can start adding and moving them over to the other side.
Well, I recognize that this is the same variables as this, so I can add these two quantities together. Here I have minus 1. If it’s not written there, we know explicitly that it’s 1. Here’s plus one-half. So minus 1 plus one-half is minus one-half. So this is also equal to minus one-half at2.
Now I’m going to move this over to the other side and end up with having canceled the terms on the left. The other step I can do now is divide by t. Of course I have to divide the whole side, right and left, by t. That cancels out that variable and so then on the right I have v now is equal to one-half at2 plus x, all divided by t.
Now I’m ready to actually plug in numbers. Let me just flip the order of that so it more naturally reads left to right. v is equal to one-half at2, that’s fine to switch this order, divided by t. So this equation is now mathematically consistent with our initial equations but now we have used the mathematics to convert it into a form that gives us our answer directly.
The v that we’re after now is equal to the final position. I have to add the product of one-half acceleration times time squared and then divide that sum by the time and that number will give me my final velocity. Let’s plug in the numbers and see what we get. Our final x-coordinate was plus 120 meters and then I have to add one-half times a, 1.8 meters per second squared, times our time squared. That was 4.1 seconds, and then I have to divide this by the time, 4.1 seconds.
Let me look at the units here real quick before I punch that into my calculator. We learned before the second squared here cancels with the 2 seconds I have in the numerator there and I end up with meters being added to meters, good sign. I’m going to end up with meters in the numerator divided by seconds in the denominator, a second excellent sign. v does have units of meters per second. So far so good. Let’s get a number for it, see if it makes sense.
I end up with the final velocity as being plus 32.96 meters per second. I would round that to three digits. 6 rounds us up and I would write 33.0 meters per second. That is how I would type it in.
Let’s see if this physically makes sense. Let me go back up here and just grab our last step because that’s what we’re doing now. Just a quick check. We’ve already checked the units, so our algebra is good. We might have plugged it into our calculator incorrectly. Let’s see if it physically makes sense.
Rule of thumb, to get from meters per second into miles per hour, which is more intuitive to us, you just roughly multiply by 2. So 33 meters per second is roughly 66 miles per hour. Let’s think about what the initial velocity was. Oh, you know what? We never got a number for that so I just don’t know.
We could go back if we wanted. Here, for example, we could use this equation, plug in our 33 meters per second using the a and the t. Actually, I would plug it into this equation. The initial velocity would be 33 minus 1.8 times 4.1 seconds. In fact, I’m going to do that calculation right now in real-time for you. 4.1, multiply these together and then I’m going to . . . I get 25.
So the initial velocity was about 26 meters per second, so that’s about 50 miles per hour. So this car appears to be going 50 when it passes the first marker and with this acceleration of roughly 1.8 meters per second squared, by the time they cross the second marker they’re going about 66 miles per hour. Seems reasonable. That’s as much as I would check to see if the answer seemed reasonable. So I’ll just put a little check mark there. Seems okay.