the-position-of-a-car-in-a-drag-race-is-measured-each-second-and-a-the-results-are-tabulated-below-b

Question

The position of a car in a drag race is measured each second, and A the results are tabulated below.
$$\begin{array}{|l|c|c|c|c|c|c|}
\hline 	ext { Time } t(\mathrm{s}) & 0 & 1 & 2 & 3 & 4 & 5 \
\hline 	ext { Position } x(\mathrm{m}) & 0 & 1.7 & 6.2 & 17 & 24 & 40 \
\hline
\end{array}$$
Assuming the acceleration is approximately constant, plot position versus a quantity that should make the graph a straight line. Fit a line to the data, and from it determine the approximate acceleration.

Robert Zaballa · Accepted Answer

If you plot the position versus the square of the time, you will indeed get the following. You can plot the points versus the, uh on a graph and do a best fit, and you will get a slope of about 1.5765 And if you double that float, you can indeed get the acceleration. So these were the data points, which I have written out here, and I used a computer in my case. You can do this by hand if you know the formulas for best fit, and you can use graph paper and you do get a best fit line with this slope and really this equation and you do get the acceleration as twice the slope and I&#x27;ll show why it&#x27;s twice the slope in just a second. But it&#x27;s a pretty good fit. The points don&#x27;t scatter too much, and you see how nicely in scrolling up and down. We can just kind of look at this that the points lie really close to the best fit line all the way to zero, and the horizontal axis is time squared in seconds. Units of second squared vertical axis is distance and meters now notice that there is a small why intercept of about 0.365 and that can just be attributed probably to error. There&#x27;s a small degree of air in the data now. One might wonderful. Why would you have the square of time as the horizontal axis? And why would you have, um, twice the slope as your acceleration? But you have to think about position as a function of acceleration as well as time. And if you right X of T, which is position is a function of time. You have the following. You have position the initial position and let&#x27;s assume we have zero initial velocity so we don&#x27;t have that germ. And let&#x27;s just say that we have an initial position and we have an acceleration term which has the square of the time. If we make t squared are independent variable, then we would have a linear plot. This would then be our slope 1/2 times acceleration. This would be our slope T squared. Would equal are independent. Variable and X not would equal our y intercept. So it&#x27;s possible that the object was not exactly at the origin when the data was taken or perhaps the measurements were not performed. If the object has a certain length, perhaps the measurements were not performed at a certain point along the objects length. And so therefore the Y intercept is not zero, even though the position is recorded as being at zero. So there&#x27;s some possible air associated with that, so this could be attributed somewhere to some error related to that, so that could account for the whiner set not being zero. However, it is small. Now when we look at the slope, it&#x27;s 1.56 or rather 1.5765 So if we take that slope and we write it down right here, 1.5765 and we said it equal to 1/2 a. This is why the acceleration will be twice the slope. We multiply both sides by two and we get an acceleration equal to twice the slope now, and the acceleration then becomes three points 153 Now the units need to be written in terms of the quantities that we plotted. So it&#x27;s always the vertical access units over the horizontal axis units. So we will right the vertical units is meters divided by the horizontally access units which your second squared. And indeed we do get acceleration units, second squared, go on the bottom. So we do have an acceleration of three 0.15 We can just go to two decimal places if we like, and just a 3.15 meters per second squared. And if you do this by hand, although it would be tedious, I would suggest a calculator or computer. You can use the following equations for the best fit line and those air given by Well, first of all, you need the mean of all your data Points X mean equals the sum over the points where you have in data points. In our case, we have six and you divide this some over in. You have the some or rather the mean of your wide data points the vertical axis. In our case, we would have in equal six. We would identify the vertical axis with X, which is positioned Position X, and we would identify Let me take that bar off. I mean, the bar just means averaging. So this is why bar would be the average over the positions. X would be identified with T squared Times Square, basically and X bar would be the average over the time squares, basically the times where values and again any in is equal to six. To calculate the slope of the best fit line, you would have this complicated quantity, which can be quite tedious to calculate. This is why you want to use a computer or calculator. You take each data point of Times Square, which would be X and subtract the mean from it. But you&#x27;re doing this in a some and you multiply these by each position minus their mean values and divide by the following quantity, which is each time squared minus their means but squared, their quantity squared. And this some appears divided by this. Some down here in On top of that, you need their Y intercept, which is given by be equals. Why average minus in which we just had up here times their X average. So this is the equation for a line. Rather, that&#x27;s why intercepted the best fit line, I should say, then the the line itself, I would just be why equals in X plus Capital B This is the equation of the line that you want, But you identify X. If you wanted to do this problem with this method, you would identify X with t squared. And why with position and I am with this the M that you see in this equation of the line is identified here with him. This capital B is identified with The camp will be here, etcetera. So it&#x27;s a very tedious process, but by hand, that&#x27;s how you would calculate the best fit. And if you do this by hand, which I did for comparison, you&#x27;ll find that the slope in is equal to 748 0.583 over 474 0.583 which then equals 1.5765 and that capital B Y intercept is equal 2.365 So now if we look and we compare these with what we obtained using a computer in this graph, we see that they are indeed an agreement with the plot from the computer

Carson Merrill · Answer

So for this problem are being asked to determine when the car speeding up and when is slowing down based on this graph. Now, be careful. We have to realize that this is the position graph. So obviously, if this was a velocity graph, we could say, Oh, the car is always speeding up, right? But we need to actually look at the position graph which is given to us right here. So velocity is going to tell us if the car is speeding up or slowing down. And we know velocity is based on the slope of the tangent line to the curve. So if you have a craft that looks like this, obviously you have a really steep slope here, which means the velocity is very high. And then here you have a less steep slope, which means it&#x27;s, um, not going as it&#x27;s not speeding up as much. This would be it&#x27;s starting to decelerate because the velocity is zero and this is meaning that the cars actually decelerating. So I&#x27;m just looking at something like that and comparing it to this. We see that it&#x27;s very steep, and it&#x27;s the car is clearly speeding up over the course of this time right here. So, um, going from about zero to queen five, we see the car is speeding up, so if you want to do ah sp for speeding the car would be speeding up from about zero. We&#x27;re starting at zero and going Teoh 0.5. So it&#x27;s in, the car is speeding up, and then we also see that it speeds up right around. Um, probably right around this 1.25 mark right here. It starts speeding up right there. Um, so that would be from 1.25 to 2. And then we see where it really slowing down is right about in this zone. Right here. We see it sped up. It reached this point right here. And then the the slope of the line is slowly decreasing. They just keep decreasing, decreasing, decreasing until it reaches this point and then it starts increasing again. But this decreasing in the slope tells us that, um, do sl for slowing down. Um, that from 0.5 to 1.25 it appears that the car is slowing down based on the position graph and then in part B were being asked at approximately what time The cars traveling the fastest on the slowest eso, obviously, right before the car starts to slow down is when it&#x27;s traveling the fastest because the velocity is increasing. That&#x27;s the car speeding up this whole time and then it starts slowing down, Right? Obviously, if we just try to take this as a velocity graph, it would look like, Oh, this is the fastest point. But we have to rise. This is the position graph. So right here is actually when the cars traveling the fastest, so do F for fastest. Um, and we realize this occurs AT T equals 0.5. Um, but based on the context of this problem, which is the amount of time after the amount of hours after five PM, we know that it&#x27;s actually going to be 5 30 that the car reaches its fastest point as far as the slowest time for the car. Um, this is going to occur right around this 1.25 mark, because this is when the car is starting to speed up again after it had slowed down as much as it did. Um, so that occurs at T equals 1.25 and in the context of the problem, once again, because it&#x27;s the amount of hours after five o&#x27;clock, this is going to be six 15.

Ze-Han Lee · Answer

right. So in this problem ah were given Ah, the relation of the velocity of the car versus time. All right. Some party I want to describe. What happened to the car? A timetable to zero. So as we can see, that T equals zero. Ah, border. Yeah, at table 20 would see that the velocity begin decreasing. Right. So take a tickle zero. So the car start digs that already. Okay, so cars started extent awaiting a teak with one and part B. How does the cars Average velocity is in. Ah, tickle. What? He was zero to teach with one competitive average velocity between t equal one and two for five. So from t equal. Ah, so part B from T equals zero to tickle one. Because he&#x27;s a receipt that every velocities over here. Right? So it&#x27;s 10 meters per second. Be bars 10 meters with second from 0 to 1 and from 1 to 5. So again we see that it&#x27;s right here, right? It&#x27;s just the middle off the highest position. Lowest position, which is over here. So it&#x27;s also 10 leave our equal two meters per second. So this is from 1 to 5. So in that sense, the average of lost city off this two individuals of the same. And Parsi says that what is the displacement of the car from 0 to 7? So Ah, because we basically have three stages. The first stages from zero want Ray. So the displacement from 01 that say it&#x27;s x 01 equal, uh, half so t is from 01 race. So it&#x27;s one and times Ah, the ah. So basically, we&#x27;re just calculating the country in the area of this triangle, right? So times 20. So this gives it two meters, and then the second stage is from 1 to 5. So when the culture of the area of this triangle okay, x 15 equal half times so five minus one and times 20. All right, so this gives you 40 meters in. Ah, there is from 5 to 7 when the calculated this area. All right, so x 57 a quote because, as you can see, that the velocity is negative in this area, right? So the area here is also negative, so it equal half negative half, and the time is from 75 to 77 line is fine and times 20. Sorry, snow. Twenties 10. Sorry. Here. Times 10. So this gives you Ah, you see, Negative 10 meters. So totally displacement will be equal to the summation of this three. All right, X equal XO one plus x 15 plus x +57 So this equal 40 meters. All right. And, ah, in the next part. Ah, we want to pluck the car&#x27;s acceleration during the interval as a function of time. Ah, hold on. I&#x27;ll use another one. Another blank paper. Upland diagram. So it&#x27;s right here. So, uh Okay, So full acceleration. I&#x27;ll use the red color. So as you can see that too. The acceleration from 01 is 20 meters. So a ah a one equal 20 meters second squared. Right. So And it is counsel int in this interval. So let&#x27;s say this is a This is 20. This is a 20 and from 1 to 5, a 15 this equal. Ah, 20 over five minus want Ray. So this equal five meters per seconds quit. So this is also a constant but a different matter. Yeah, And from flying to seven. We see that accelerations negative right, A 57 equal. Negative. 10 over seven, minus five, Which is negative. Five meters seconds. Quiet, Ray. So it&#x27;s It&#x27;s like this. All right, so this is X generation. His Parsi. Uh, let me double check the Parsi a party. Oh, it&#x27;s part DEA. There&#x27;s a party, and then, ah, we want to draw. We want to make a sketch off the position during the trouble. Ah, as a function of time and assume that the initial position zero. So we used in another color. Ah. So basically see that from 01 it is accelerating. Right? So the ah, the movement is like is like, um, problem. And it is open up. So, like this, right? And from one to five, as we can see, the velocity is also it It&#x27;s ah, it is positive, but it is the thinks that already so the displacement will be like this at the Tiegel five richest the Maxima. Okay. And, uh, from 5 to 7, state velocity becomes negative rain, so the displacement should decrease in this interval. So you mean like this? So this is the diagram off the displacement

Averell Hause · Answer

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&#x27;s what&#x27;s giving us this. Ah, great cur. There&#x27;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&#x27;re simply using the, uh, formula here, and I&#x27;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&#x27;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&#x27;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&#x27;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&#x27;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&#x27;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

Josh Broderick Phillips · Answer

for this problem were given position and time data for ah, race car and we want to find the velocity as a function of time. In the exploration is a function of time and the way I&#x27;m going to do that, I&#x27;m just gonna use numerical differentiation and excel to make this really quick. So if we want to find a given velocity to do Delta X over Delta T, we take the position one at 11 greater position and one less position that interval and divided by the changing time over that same interval. And we do that all the way down for each value until we get to hear and for the last one, we don&#x27;t have anything in these spots. So every time you do numerical differentiation, you&#x27;re going to lose the edge values. Except we also know that the race car starts at rest. That&#x27;s what we&#x27;re given in the problem. So I just plugged in zero there. Then we can do the same thing for the acceleration. Just take the difference of the velocity and divided by the time difference and do that all the way down. And now we get our approximate values for velocity and exploration at each time and that if we want to graft these, this is what it looks like. The velocity. It&#x27;s pretty linear, Um, of you. It&#x27;s hard to say exactly what it is, but it looks linear from the broad perspective, and I want to be getting to acceleration and we can see it. Actually, it&#x27;s the subtle changes in velocity or magnified. Um, but what I&#x27;m doing for these graph is just graphing these columns. So velocity versus time, acceleration versus time. I&#x27;m just graphing those values, and that&#x27;s what I that&#x27;s what it looks like.

Zulfiqar Ali · Answer

well from the first part of the motion we have, Beban is equal to 0.60 square and then we have we I V I is equal to zero. Ex relation in this part is equal to everyone And everyone is equal to d Rayvon divided by DT which is equal to 1.2 times T and distance travelled is equal to integral off DS physical to integral off We want times tt so here yes, one is equal to integral from zero to be see to find six t square DT So if you see lines, it s one physical too. See the point to multiply by Take you right No x relation eight at the end of this part when time rt physical too wipe Second is ah is equal to we have ah 1.2 one point to multiply by five multiply by five which is equal to six o meter per second square by distance traveled after fights against these s is equal to 0.2 multiplied by 125 That is equal to 25 meter. Now a gradient of the second part of motion is Are we to minus 15 is equal to zero minus 15. Divided by 15 minus five into D minus five Right, And further we do is a cold too. Minus 1.5. Do you? Let&#x27;s 22.5 22 Point life No explanation of the second Bart. Exploration of the second part is equal to on TV two divided by DT Right, So that is equal to minus 1.5. Me too. Per second squared now distance traveled is given by integral off ts of limits are the lower limit is 25 licked upper limit bs to physical to federal off we to gaity next. Supplying the mixed five to T and sold in guessing digger list too is equal to minus is defined 75 the square less 22.5 22.5 D minus 68 Wine toe 75 75 Right now let&#x27;s ah dro, Let&#x27;s plot the grounds. Well, first of for will be drawn ex relation time grab right. So Flextronics relation horses down. So here is the X relation. And it was the time that this 0.0 Series five um then on 15 right? You scroll down a little bit. Okay, so the drop would look like this. So the axe relation is increasing. Right? So till five seconds. Right. And then these negative and constant from 5 to 15 seconds. I&#x27;m so it would look like this. And now let&#x27;s draw position versus time graph. Let me right here. He sees time. And here is the position. This is a region zero. Why, uh, disease 10 and 15. So here is 50 100 and so on. So the position horses time drop would look like this. So it is the position horses, dandruff.

Problem

A fireworks rocket explodes at a height of $82.0 …

Problem 52 Hard Difficulty

Answer

$$3.153 \mathrm{m} / \mathrm{s}^{2}$$

Related Courses

Essential University Physics

Related Topics

Discussion

Top Physics 101 Mechanics Educators

Liev B.

Marshall S.

Zachary M.

Meghan M.

Physics 101 Mechanics Courses

Math Review - Intro

Algebra - Example 1

Recommended Videos

Watch More Solved Questions in Chapter 2

Video Transcript

Richard Wolfson

Essential University Physics

Liev B.

Marshall S.

Zachary M.

Meghan M.

Math Review - Intro

Algebra - Example 1

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$$3.153 \mathrm{m} / \mathrm{s}^{2}$$

Essential University Physics

Liev B.

Marshall S.

Zachary M.

Meghan M.

Math Review - Intro

Algebra - Example 1

Richard WolfsonEssential University Physics

Liev B.

Marshall S.

Zachary M.

Meghan M.

Richard Wolfson

Essential University Physics