Hey, guys. Another type of graph that you'll need to solve motion problems is a velocity-time graph, and we can use these graphs to solve things like acceleration. Now, guys, this is going to be exactly like how we use position-time graphs for velocity. So we're going to be reusing a lot of the same concepts that we've already discussed, but just the variable is going to change. That's really all there is to it. So let's check it out. With a position-time graph, remember that these showed the position on the y-axis versus time on the x-axis. Well, the most obvious difference between these new velocity-time graphs is that the y-axis shows velocity instead and the x-axis still shows time. So, really, the only thing that's different is just what the y-axis represents.

Now moving on, we saw that the velocity, represented by ΔxΔt, was the slope of our position graph. Here with velocity-time graphs, it's the same principle. We're going to use the slope, except now it's just going to represent the acceleration. So the acceleration is the slope of the velocity-time graph. When we talk about position-time graphs, we discussed two different kinds of velocities or slopes. The slope between two points was an average velocity and the slope of the tangent line at one point was called the instantaneous velocity. For example, let's say I wanted the slope between these two points over here, draw the line, and then this represents the average velocity. Whereas, if I wanted the velocity at a particular point, let's say at t4, you'd have to draw the tangent line which again was kind of our best guess here, and this tangent line was just the instantaneous velocity.

It's the exact same thing over here. If I want the slope between these two points, this is going to represent an average acceleration now. So this is just the letter that's different. And then if I wanted the acceleration at a particular point, then you just draw the tangent line. So this tangent line here is going to represent or the slope of the tangent line is going to represent the instantaneous acceleration.

The last point is just the steepness of the slopes and what they represent. In position-time graphs, if you had steeper slopes, for instance, if you had slopes that got more and more vertical, then what that represented is that your velocity was higher in magnitude. Again, higher just means the number was bigger. Forget about the sign. So more vertical meant that the numbers were bigger. Well, it's the same idea here. A steeper slope on a velocity-time graph just represents now a higher magnitude acceleration. So the more vertical one line gets, then that means that the acceleration, again the number, forget about the sign, just gets bigger. Alright, guys. So you'll see that we're using a lot of the same ideas and a lot of the same concepts.

So let's just get to an example and see how this works. So we've got the velocity-time graph for a moving car over here, and we're going to figure out the acceleration in this first part between t15−25. So let's just get to it. So the first thing is that we're solving for acceleration, but we're using two different times. So which one are we going to use? Are we going to use a tangent line, or are we going to use an average? Well, again, this is between two points over here. So this is going to be an average acceleration. We're going to draw the line that connects t15−25 which is from here to here, and we actually don't have to draw a line because it's already drawn for us. So we just have to figure out the slope of this line. So I can make a little triangle like this and I figure out my rise over my run. So this is actually going to be Δv over Δt, not Δx, and so, let's see. So my Δv is going to be, well, I'm going all the way down to 0 actually. I'm going to end up at 0 over here. So this is going to be 0 minus and this is going to be 60. So that means that my Δv is actually negative 60. And the time is from 15 to 25. So my Δt is 25−15, and that's 10. So that means that my average acceleration is negative 6 meters per second squared. So this is my average acceleration. It makes sense because the slope is downward. All those same principles still apply here in acceleration.

So let's move on to now the acceleration at T10. So now, we're asked for the acceleration at one particular point, not between two. So that means that we're looking for the instantaneous acceleration when t10 is equal to 10. First, we have to figure out the tangent line at this point. The tangent line here is going to be if I trace along this graph and then, instead of following the curvature, I kind of just keep on going in a straight line. So my best guess for this tangent line looks kind of like, I don't know, something like this over here. So now, this instantaneous velocity is going to be the slope of this tangent line or at least my best guess at it. So I still am going to need a Δv over Δt. Remember, I still need to calculate the rise over the run, so I need two points. Except now I'm going to use these two points over here at the endpoints. My final velocity is going to be 75. My initial velocity is going to be 30. So my Δv is going to be 75 minus 30, which is equal to 45, and so that I know this is 45, and the Δt is going to be from 5 to 15 seconds. So that means that my Δt is just 10 seconds. So that means 45 over 10 is 4.5 meters per second squared, and it's positive. So that means the slope of this line, which is the instantaneous acceleration, is about 4 and a half meters per second squared. Again, kind of just a best guess. Alright, guys. That's it for this one. Let me know if you have any questions.