## Video transcript

Let's work through anotherfew scenarios involving displacement, velocity, andtime, or distance, rate, and time. So over here we have, Ben isrunning at a constant velocity of three 3 meters persecond to the east. And just as a review,this is a vector quantity. They're giving us themagnitude and the direction. If they just said 3meters per second, then that would just be speed. So this is the magnitude,is 3 meters per second. And it is to the east. So they are givingus the direction. So this is a vector quantity. And that's why it'svelocity instead of speed. How long will it takehim to travel 720 meters. So let's just remindourselves a few things. And I'll do it both withthe vector version of it. And maybe they shouldsay, how long will it take them to travel720 meters to the east, to make sure, to make it clear,that it is a vector quantity. So that it's displacement,as opposed to just distance, but we'll do it both ways. So one way to think about it, ifwe think about just the scalar version of it, we saidalready that rate or speed is equal to the distance thatyou travel over some time. I might write t there. But it's reallya change in time. So sometimes some peoplewould write a little triangle, a delta there, whichmeans change in time. But that's implicitlymeant when you just write over time like that. So rate or speed is equalto distance divided by time. Now, if you know-- they'regiving us in this problem, they're giving us the rate. If we think about thescalar part of it, they're telling us thatthat is 3 meters per second. And they're alsotelling us the time. Or, sorry, they're nottelling us the time. They are tellingus the distance, and they want us tofigure out the time. So they tell us thedistance is 720 meters. And so we just have tofigure out the time. we So if we just do thescalar version of it, we're not dealing withvelocity and displacement. We're dealing with therate or speed and distance. So we have 3 metersper second is equal to 720 meters oversome change in time. And so we can algebraicallymanipulate this. We can multiply bothsides times time. Multiply time right over there. And then we could,if we all-- well, let's just take itone step at a time. So 3 meters per second timestime is equal to 720 meters because the times on theright will cancel out right over there. And that makes sense,at least units-wise, because time is goingto be in seconds, seconds cancel out theseconds in the denominator, so you'll just get meters. So that just makes sense there. So if you want tosolve for time, you can divide both sidesby 3 meters per second. And then the leftside, they cancel out. On the right handside, this is going to be equal to 720divided by 3 times meters. That's meters in the numerator. And you had meters persecond in the denominator. If you bring it outto the numerator, you take the inverse of this. So that's meters-- let me dothe meters that was on top, let me do that in green. Let me color code it. So 720 meters. And now you're dividingby meters per second. That's the samething as multiplying by the inverse, timesseconds per meters. And so what you'regoing to get here, the meters aregoing to cancel out, and you'll get 720divided by 3 seconds. So what is that? 720 divided by 3. 72 divided by 3 is 24. So this is going to be 240. This part right overhere is going to be 240. And it's going tobe 240 seconds. That's the only unitwe're left with, and on the left hand side,we just had the time. So the time is 240 seconds. Sometimes you'll see it. And just to show you,in some physics classes, they'll show youall these formulas. But one thing I reallywant you to understand as we go through thisjourney together, is that all of thoseformulas are really just algebraicmanipulations of each other. So you really shouldn'tmemorize any of them. You should alwayssay, hey, that's just manipulating one of those otherformulas that I got before. And even these formulas are,hopefully, reasonably common sense. And so you can start fromvery common sense things-- rate is distancedivided by time-- and then justmanipulate it to get other hopefullycommon sense things. So we could have done it here. So we could have multipliedboth sides by time before we even putin the variables, and you would have gotten--So if you multiplied both sides by time here,you would have got, on the right hand side, distanceis equal to time times rate, or rate times time. And this is oneof-- you'll often see this as kind ofthe formula for rate, or the formula for motion. So if we flip it around, you getdistance is equal to rate times time. So these are allsaying the same things. And then if you wantedto solve for time, you could divideboth sides by rate, and you get distance dividedby rate is equal to time. And that's exactly what we got. Distance divided byrate was equal to time. So if your distanceis 720 meters, your rate is 3meters per second, 720 meters divided by3 meters per second will also give you atime of 240 seconds. If we wanted to dothe exact same thing, but the vector versionof it, just the notation will look a littlebit different. And we want to keep trackof the actual direction. So we could say weknow that velocity-- and it is a vector quantity,so I put a little arrow on top. Velocity is the samething as displacement. Let me pick a nice colorfor displacement-- blue. As displacement-- Now, remember,we use s for displacement. We don't want touse d because when you start doing calculus,especially vector calculus-- well, anytype of calculus-- you use d for thederivative operator. If you don't know what that is,don't worry about it right now. But this right here,s is displacement. At least this is convention. You could kind of useanything, but this is what most people use. So if you don't wantto get confused, or if you don't want to beconfused when they use s, it's good to practice with it. So it's thedisplacement per time. So it's displacementdivided by time. Sometimes, once again,you'll have displacement per change in time,which is really a little bit more correct. But I'll just go withthe time right here because this is the conventionthat you see, at least in most beginning physics books. So once again, if wewant to solve for time, you can multiplyboth sides by time. And you get-- this cancelsout-- and I'll flip this around. Well, actually, I'llleave it like this. So you get displacementis equal to-- I can flip these around--velocity times change in time, I should say. Or we could just say timejust to keep things simple. And if you wantto solve for time, you divide bothsides by velocity. And then that gives you timeis equal to displacement divided by velocity. And so we can apply thatto this right over here. Our displacement is720 meters to the east. So in this case, our timeis equal to 720 meters to the east. 720 meters east isour displacement, and we want to divide thatby the given velocity. Well, they give us thevelocity of 3 meters per second per the east. And once again, 720 dividedby 3 will give you 240. And then when you takemeters in the numerator, and you divide bymeters per second in the denominator, that'sthe same thing as multiplying by seconds per meter,those cancel out. And you are just leftwith seconds here. One note I want to give you. In the last few problems, I'vebeen making vector quantities by saying to theeast, or going north. And what you'regoing to see as we go into more complex problems--and this is what you might see in typical physicsclasses, or typical books, is that you define a convention. That maybe you'll say, thepositive direction, especially when we're just dealingwith one dimension, whether you can eithergo forward or backwards, or left or right. We'll talk about othervector quantities when we can move in twoor three dimensions. But they might takesome convention, like positive means maybeyou're moving to the east, and maybe negative meansyou're moving to the west. And so that way-- well,in the future, we'll see, the math willproduce the results that we see a little bit better. So this would just bea positive 720 meters. This would be a positive3 meters per second. And that implicitly tellsus that that's the east. If it was negative, itwould then be to the west. Something to think about. We're going to start exploringthat a little bit more in future videos. And maybe we might say positiveis up, negative is down, or who knows. There's differentways to define it when you're dealingin one dimension.