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Recorded Position of a car moving in a straight linein _timetime (sec) X (meters)5.0 12410.0 15.0 20.0 25.0 30.0 35.0 40.0 152 175 202 225 245 275 305Plot this data...

Question

Recorded Position of a car moving in a straight linein _timetime (sec) X (meters)5.0 12410.0 15.0 20.0 25.0 30.0 35.0 40.0 152 175 202 225 245 275 305Plot this data on a graph as x Vs. L and determine the slope of the line obtained as best as You can_ Show this slope calculation on the graph: The slope of the line represents the rate at which the distance changes with time; i.e,, the average speed of the car: Determine the average speed of the car and estimate the uncertainty in this speed (that

Recorded Position of a car moving in a straight linein _time time (sec) X (meters) 5.0 124 10.0 15.0 20.0 25.0 30.0 35.0 40.0 152 175 202 225 245 275 305 Plot this data on a graph as x Vs. L and determine the slope of the line obtained as best as You can_ Show this slope calculation on the graph: The slope of the line represents the rate at which the distance changes with time; i.e,, the average speed of the car: Determine the average speed of the car and estimate the uncertainty in this speed (that is, estimate the precision of the slope)_ Write down your result here for the average speed. Include the proper units. Vayerage '



Answers

A model car starts from rest and travels in a straight line. A smartphone mounted on the car has an app that transmits the magnitude of the car's acceleration (measured by an accelerometer) every second. The results are given in the table:
Each measured value has some experimental error. (a) Plot acceleration versus time and find the equation for the straight line that gives the best fit to the data. (b) Use the equation for $a(t)$ that you found in part (a) to calculate $v(t)$, the speed of the car as a function of time. Sketch the graph of $v$ versus $t$. Is this graph a straight line? (c) Use your result from part (b) to calculate the speed of the car at $t =$ 5.00 s. (d) Calculate the distance the car travels between $t =$ 0 and $t =$ 5.00 s.

Alright, we're given an equation for X versus T. So I'm not sure I could read this correctly. X. Is five T Plus 0.75 T. Cute. Okay. A plot the position versus time. Oh man, I wish I had showing the dez most calculator here. Um I'm gonna graph this in dez mose Y equals five T plus 0.75 T. Cute. And then I'm going to just draw it. It looks like an elongated cube function. So it's kind of like this only it's really more stretched upward like that. I mean the fact of the matter is if I were to elongated oops, it would be Uh huh. More like like that. Okay. Um do do do do do do do do do instantaneous velocity at four seconds. Well, if I take the derivative of this velocity is five plus .75 times three .75 times three It's 2.25 times T squared. So v at four is going to be five plus 2.25 times 16 41 meters per second. See Average velocity in four seconds. While average velocity is going to be distance at four over four Acts of four is going to be five times 4. Putting it into the axe equation plus 0.75 times for cubed. Which is 68, 68 divided by four is 17. All right, so the instantaneous velocity is more than twice the average velocity. Thank you for watching

So this would be the, uh, plot of the the position versus time graph. And we see the parabolic function here. This is Ah, we have, ah, function of time, race to the third power. And so that's what's giving us this. Ah, great cur. There's a exponential curve here. And so we can say that four part Ah, a continuing on for party. We can say that. Um well, these points are simply found at one second to two seconds and three seconds. And then you can simply connect the points in order to sketch the graph eso again, just with the function equaling except T equaling 5.0 meters per second multiplied by tea. And this would be a plus 0.75 meters per second cube multiplied by t cubed Ah, you can simply plug in 12 and three Ah, for tea and find the these points for one. It should be. This would be 5.75 here. This should be 16. And here this should be 35 0.25 So, after plug in, you should get those values and then you can simply sketch for part B now and wants you to find the instantaneous velocity. This would be at T meets a instantaneous velocity at T equaling 4.0 seconds were first using time intervals of 0.4 seconds. So we can say that. Ah, we can calculate ex at T equaling 3.8 seconds and ex at T equaling 4.2 seconds. Again, we're simply using the, uh, formula here, and I'm gonna lose the units so we can. Ah, so it could be a little bit quicker too. Calculate. So five times 3.8, this would be plus 0.75 times 3.8 Cute and then 44.2 seconds. It would be five times 4.2 plus 0.75 times 4.2. Again cute. And so, finding the instantaneous velocity, it would simply be the difference between the the change of displacement with respect to time. So this would be equaling 76.6 meters minus 60.15 meters, and then this would be divided. My 0.4 seconds. And so the instantaneous velocity is gonna be equaling approximately 41.1 meters per seconds. Again, AT T equals 4.0 seconds. Now we're going to be doing the exact same thing. So this would be your answer for a time. Interval of 0.4 seconds. Now we're gonna do the exact same thing. However, our time intervals now 0.2 seconds. So here we have ex at T equaling 3.9 seconds. I'm an ex at T equaling 4.1 seconds. So this is gonna be equaling again. The exact same thing. Five times 3.9 plus point 75 times 3.9. Cubed again. Uh, brother, five times 4.1 plus 0.75 times 4.1 quantity cubed the instantaneous velocity. Is that gonna be equaling 72.2 meters minus 64 meters. I get it simply All right. T X of tea at four point x of tea equaling 4.1 seconds minus x of tea equaling 3.9 seconds. So is simply the difference. And then here, of course, before 0.1 seconds, minus 3.9 seconds. Of course, this is equaling 0.2 seconds, and so this is equaling 41 meters per second. So this would be your answer for part B for a time interval of again. Ah, 0.2 seconds. So delta t equal in 0.2 seconds. Now finally, point. Uh, we have 0.0.1 2nd And so this is where it gets a bit more accurate. Where? Ex at t equaling 3.95 seconds and then ex at T equaling 4.5 seconds. So this would be equaling again. Five times 3.95 plus 0.75 times 3.95 toothy third power. All right. 4.5 plus 0.75 times 4.5 to the third power. And the instantaneous velocity is equaling here. 70.1 minus 66.0. Divided by here. Ah, 4.5 minus 3.5 And this is equaling again. 41 meters per second. So this would be the instantaneous velocity. At four seconds, we're getting closer and closer. So 41 we use per second is essentially pretty close to the exact answer. And so we can say that for part C to find the average velocity, this would be equaling ex At T equaling 4.0 seconds minus x AT T equaling zero seconds and then this would be divided by 4.0 seconds, minus zero seconds. And so this is equaling five times four plus 0.75 times four cube. This would be divided by four. And this is equaling 17 meters per second. So you can see that this average velocity is much less. 17 meters per second is much less than 41 meters per second. And so the average velocity Ah, throughout the entire time interval of 4.0 seconds, or rather from zero seconds to four seconds. This is gonna be ah lot less than the instantaneous velocity at four seconds. So, again, this is taking into account your total displacement from the beginning to the end and how long it took to get there. This is going to be the tangent line at that point in the curve. So with the parabolic functions such as this one, of course, we're going to get a higher, instantaneous velocity. The greater the more time has passed, then if we calculate the entire average velocity from the beginning, So again, the average velocity from zero seconds to four seconds is much much Wes the instantaneous velocity at four seconds, that is the end of the solution. Thank you for watching

Using the graph given here of the velocity of two cars. And we want to find in part a how far car has traveled after two seconds. And so the idea there is were given. Here's two seconds were given feet her second, and the general shape when we take the integral is a rectangle. So let's just go ahead and draw a rectangle. If we're going to find the area under the curve, well, the height would be feet per second, and then we'd multiply by the with, which is seconds. But if we multiplied those two, we just end up canceling the seconds and get feet. That's why taking the integral tells us the distances because those units cancel like that and so doing so we could go ahead and split this up into a triangle. And that's triangle. Here is one half base times height on the bases, one the height of six, and so that would have an area of three. And then this rectangle is just base times, height and that would be one time six. The oh yeah, wait one time six. That's right. So that would be six. Therefore, both of those added together, those two shapes would give us this 9 ft. Okay. Part B one, his car a farthest ahead of Karbi. Well, car has a has a greater velocity this whole time all the way up until two seconds. And so that whole time it Z speed is greater. Therefore it's going to be further away. But then at this point, Carby starts toe catch up and then over well, catch up to car. So at two seconds is basically where Car A was the furthest ahead of Karbi for part C. We've now want to estimate that black area that I just shade shaded, which is how far Carrey is ahead of Karbi. Now we already know that we have the area under the curve of a from 0 to 2 and that is 9 ft. What we now want to do is calculate the area under the curve of Karbi, which is all of this area, and subtract that out. And then that subtracted out. We'll end up giving us that black area. So that's the idea. But we're gonna estimate that area using a bunch of rectangles. They each have a width of one half and so we're going to use for them. So let's just go ahead and put each of these rectangles here the first one. Sorry. I know it's hard to see them use. Green is like right here. For instance, the next one is, like right here, them here and then here. And we're using the midpoint for each of those rectangles. Estimate the area for each. So first we've got each of our widths one half all the way across. And then finally, we have the fourth one there. Now let's go back. And let's estimate the heights for these and that would end up giving us this'll. One. Looks like a height of 10.25 When we look at the midpoint there, the next one. Sorry, it looks like it has a height of about one. The next one has a height in between one and 1.5. Hmm. Maybe about 2.5. So that height there for the 3rd 12.5. And then finally, the third one, maybe about five. What could say? So that height here would be five. Now, let's just go ahead and put these in real fast, so we're going to do. I'm just gonna type this end to the calculator here. 0.5. I'm to a 0.25 and then plus the next rectangle will that be one time? 0.5. So if you just leave that plus 2.5 times 0.5 and then plus ah, five times 0.5 or just 2.5. But I'll just type it out anyway. Okay? Enter. And that gets us about 4.4 for these areas under the blue curve. And that's feet. So that's how far the blue car traveled. And therefore, if we do 9 ft and then subtract the 4.4 ft, then that will give us, um, What is that? 3.6 or 4.6? Yeah, 4.6. That's That's all right. Mental math. Not always my strongest, So nine minus that 4.6 it ahead. Okay, so I hope that help clarify what's going on with that one. And then, lastly, we have Part D. Let's go back to the graph. And so D is saying, give a rough estimate of one car be catches up with car A. I'm going to redraw the graph really quickly because it's just kind of a mess. We wanna I lights, um, important things. So we just calculated this is the distance that car A was ahead of Karbi. We want to find out when that area matches perfectly over here, and we want to then go down to what time that happens. Well, it's hard to tell based on this graph, but it looks like it's gonna happen at least between three times three and time is 3.5 somewhere in between. That gap is when the the area of part B is going to then, um, matchup back with car A completely. So I hope that explanation made sense here. And that's the idea. What velocity graphs is integrating them gives you distance.


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