When velocity is constant, you can multiply the velocity by the time and get the displacement: <\/p>\n\n\n\n But it is interesting to\nlook at a velocity vs. time graph that models this situation:<\/p>\n\n\n\n The area between the graph and the t-axis forms a rectangle. So when you are multiplying the velocity (6 m\/s) by the time (5 s) you are also finding the area of that rectangle! <\/p>\n\n\n\n But is this really<\/em> an area? Or is it more of an “area”? Look at the lengths of the sides and note the units. We expect the lengths of the sides of a rectangle to be measured in units of distance. But this one has one side measured in velocity units (m\/s) and the other in units of time (s). And when you multiply them, you get a unit of distance, not a typical unit of area. So the “area” on the graph<\/em> represents some other physical quantity we care about.<\/p>\n\n\n\n <\/p>\n\n\n\n <\/p>\n\n\n\n Now here’s something your physics teacher might ask THIS year:<\/p>\n\n\n\n A car\u2019s velocity increases with time according to the function v(t)=t2<\/sup><\/strong>. How far does the car travel in the time from t = 0 to t = 5 s ? <\/em><\/p>\n\n\n\n Since we know how to find displacement when the velocity is constant, let’s adapt the idea we used in math class to generate an approximate answer. <\/p>\n\n\n\n Here is our plan: <\/p>\n\n\n\n So you are pretending that instead of gradually increasing, the velocity stayed constant for 1 second at a time, jumping to the next value at the beginning of each second (depending on whether we want to over-estimate or under estimate our answer). <\/p>\n\n\n\n <\/p>\n\n\n\n Why don\u2019t we graph what thAT looks like on a velocity vs time graph?<\/p>\n\n\n\n \u201cIt\u2019s like D\u00e9j\u00e0 vu all over again\u201d \u2013 Yogi Berra<\/em><\/p>\n\n\n\n I hope this looks familiar! It\u2019s just like what we saw in the last post. Treating the motion as if if were a series of constant-velocity intervals is just like approximating the area under a curved boundary by using a bunch of rectangles!<\/p>\n\n\n\n I used a spreadsheet and put the results of these calculations in the chart below. Take a look at the overestimate and the underestimate. You might recognize those numbers…but once you understand what we are doing here, you won’t have to do these calculations at all!<\/p>\n\n\n\n <\/p>\n\n\n\n So now we can say that the cart travels somewhere between 30 and 55 meters. You can probably guess what your teacher will say:<\/p>\n\n\n\n \u201cI like your method, but I\nwish it were more precise. Can you\ndivide the interval into more, finer intervals? \nLike maybe 10 of them? Or even 50? \n<\/p>\n\n\n\n [It\u2019s amazing how your math teacher and physics teacher are so in tune. It\u2019s almost as if they were one person\u2026]<\/p>\n\n\n\n Here\u2019s what 50 rectangles\nlook like (again). Seriously, the only difference is that I labeled the x and y\naxes to be velocity and time.<\/p>\n\n\n\n And if we want the exact\nanswer, we evaluate the limit:<\/p>\n\n\n\n Except we still<\/em> don\u2019t have to do it this way! We\ncan still use the Magic Theorem! Really, all that has changed is that we are\ncalling the function v(t) instead of f(x):<\/p>\n\n\n\n It is true that we have just\nfound an area under a graph. But what we\nalso found, and what we actually wanted\nto find<\/em>, is the answer to a physics question: how far did the car\nmove? It is just a helpful connection\nthat we can think of that as an area.<\/p>\n\n\n\n IT\u2019S NOT JUST FOR DISPLACEMENT <\/em><\/strong><\/p>\n\n\n\n Here are some other things we\nwill calculate this year. In each case, we can think of it as an area and then\nuse the Magic Theorem. <\/p>\n\n\n\n 1. Impulse exerted by a force\nthat varies with time:<\/p>\n\n\n\n This is also the area under\nthe Force vs time graph.<\/p>\n\n\n\n 2. Charge delivered by a current\nthat varies with time:<\/p>\n\n\n\n This is also the area under\nthe current vs time graph.<\/p>\n\n\n\n 3. Work done by a force that\nvaries with position:<\/p>\n\n\n\n This is also the area under\nthe force vs. position graph.<\/p>\n\n\n\n Once you are comfortable with this, you can find these things without even drawing the graphs! Integrating will come to mean: adding up infinitely many tiny contributions to an overall total. And as you will see in the next (and last!) post in this series, that\u2019s another powerful way use integration in physics.<\/p>\n\n\n\n Ahead to Part V<\/a><\/p>\n\n\n\n Back to Part III<\/a><\/p>\n\n\n\n Go to Summer Reading Outline<\/a><\/p>\n","protected":false},"excerpt":{"rendered":" Here is an easy warm-up from your first year of physics class: A car moves at a constant velocity of 6 m\/s for 5 seconds. How far does it travel during that time? When velocity is constant, you can multiply … Continue reading ![\"\"](\"http:\/\/advancedmathyoungstudents.com\/blog\/wp-content\/uploads\/2021\/08\/magic10.png\")
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