Determine the relationship between speed and distance traveled over time

Speed & Velocity – The Physics Hypertextbook

In a physics equation, given a constant acceleration and the change in velocity of an object, you can figure out both the time involved and the distance traveled. Given that time in all likelihood is solely a human notion with no material existence time and speed is that speed is the amount of distance traveled over time. So speed is dependent on time, as in the speed is determined by the distance. The equation. When an object moves in a straight line at a steady speed, you can calculate its speed if you know how far it travels and how long it takes. This equation shows the relationship between speed, distance travelled and time taken.

So they tell us the distance is meters. And so we just have to figure out the time. We're dealing with the rate or speed and distance. So we have 3 meters per second is equal to meters over some change in time. And so we can algebraically manipulate this. We can multiply both sides times time. Multiply time right over there. And then we could, if we all-- well, let's just take it one step at a time. So 3 meters per second times time is equal to meters because the times on the right will cancel out right over there.

And that makes sense, at least units-wise, because time is going to be in seconds, seconds cancel out the seconds in the denominator, so you'll just get meters. So that just makes sense there. So if you want to solve for time, you can divide both sides by 3 meters per second. And then the left side, they cancel out. On the right hand side, this is going to be equal to divided by 3 times meters.

That's meters in the numerator. And you had meters per second in the denominator. If you bring it out to the numerator, you take the inverse of this.

Total distance traveled with derivatives

So that's meters-- let me do the meters that was on top, let me do that in green. Let me color code it. And now you're dividing by meters per second. That's the same thing as multiplying by the inverse, times seconds per meters. And so what you're going to get here, the meters are going to cancel out, and you'll get divided by 3 seconds.

So what is that? So this is going to be This part right over here is going to be And it's going to be seconds. That's the only unit we're left with, and on the left hand side, we just had the time. So the time is seconds. Sometimes you'll see it. And just to show you, in some physics classes, they'll show you all these formulas. But one thing I really want you to understand as we go through this journey together, is that all of those formulas are really just algebraic manipulations of each other.

So you really shouldn't memorize any of them. You should always say, hey, that's just manipulating one of those other formulas that I got before. And even these formulas are, hopefully, reasonably common sense. And so you can start from very common sense things-- rate is distance divided by time-- and then just manipulate it to get other hopefully common sense things.

So we could have done it here.

So we could have multiplied both sides by time before we even put in 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, distance is equal to time times rate, or rate times time. And this is one of-- you'll often see this as kind of the formula for rate, or the formula for motion. So if we flip it around, you get distance is equal to rate times time. So these are all saying the same things. And then if you wanted to solve for time, you could divide both sides by rate, and you get distance divided by rate is equal to time.

And that's exactly what we got. Distance divided by rate was equal to time. So if your distance is meters, your rate is 3 meters per second, meters divided by 3 meters per second will also give you a time of seconds. If we wanted to do the exact same thing, but the vector version of it, just the notation will look a little bit different.

And we want to keep track of the actual direction. So we could say we know that velocity-- and it is a vector quantity, so I put a little arrow on top. Velocity is the same thing as displacement. Let me pick a nice color for displacement-- blue. As displacement-- Now, remember, we use s for displacement. We don't want to use d because when you start doing calculus, especially vector calculus-- well, any type of calculus-- you use d for the derivative 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 use anything, but this is what most people use. So if you don't want to get confused, or if you don't want to be confused when they use s, it's good to practice with it. So it's the displacement per time. Even though my position right over here is going to be negative 1.

Or you could say my net distance, or you could say my displacement is negative 1. I'm 1 to the left of where I started. The total distance is 7. So now we've clarified that. I encourage you to now pause this video and try to answer the question. What is the total distance traveled by the particle in these first 6 seconds?

So the easiest way I can think of addressing this is to think about, well, when is this thing moving to the right and when is it moving to the left? And it's going to be moving to the right when the velocity is positive, and it's going to be moving to the left when the velocity is negative. So this really boils down to thinking about when is the velocity positive or negative. And to think about that, let's actually graph the velocity function or make a rough sketch of it.

So this is the position function. The velocity function is going to be the derivative of the position function with respect to time. And then we have minus 12t plus And so let's just try to graph this. This is going to be an upward opening parabola. This is clearly a quadratic. And the coefficient on the second degree term, on the t squared term, is a positive number, so it's going to be an upward opening parabola. It's going to look something like this.

And we're assuming that it switches direction. So it's going to be positive some of the time and negative for some of the time. So it should intersect the t-axis where it's negative.

The function is going to be negative in that interval, and it's going to be positive outside of that interval. So the easiest thing I could think of doing is let's try to find what the 0's are. Then we can draw this upward opening parabola. So to find its 0's, let's just set this thing equal to 0 so we get 2t squared minus 12t plus 10 is equal to 0.

Divide both sides by 2 just to get rid of this 2, make this leading coefficient a 1. We get t squared minus 6t plus 5 is equal to 0. That made it a lot easier to factor. This can be factored into t minus 1 times t minus 5. Negative 1 times negative 5 is 5. Negative 1 plus negative 5 is negative 6. This is equal to 0. So this left hand side of the equation is going to be equal to 0 if either one of these things is equal to 0.

Take the product of two things equaling 0, well, you get 0 if either one of them is 0. So either t is equal to 1 or t is equal to 5. So now let's graph it. So let's draw our axes. So I could say that's my velocity axis. And let me draw the-- we only care for positive values of time.