Kinematics graph analysis

1. Dec 21, 2013

Kyle_N22

1. The problem statement, all variables and given/known data

Here is the question..

Figure P2.35 (A link to the image is posted) represents part of the performance data of a car owned by a proud physics student. (a) Calculate the total distance traveled by computing the area under the red-brown graph line. (b) What distance does the car travel between the times t = 10s and t = 40s? (c) Draw a graph of its acceleration versus time between t = 0 and t = 50s. (d) Write an equation for x as a function of time for each phase of the motion, represented by the segments 0a, ab, bc. (e) What is the average velocity of the car between t = 0 and t = 50s?

http://www.mazzworld.net/HU2Q2Image1.GIF [Broken]

2. Relevant equations

xf = x0 + v0t + (1/2)axt2
xf = x0 + v0t

3. The attempt at a solution

For part a, I used a formula based on the geometry of the graph line simply to calculate the area under the curve which I found to be 1250m. For part b, I broke up the total area under the graph line from t = 10s to t = 40s into geometric sections, calculated the area under each section, then took their sum to find 1450m. For part c, I used the slope of the line to find the acceleration from 0a, ab, and bc. Of course, the acceleration is constant from 0a and bc, but it is 0 from ab. I used this info to obtain an appropriate graph. For part d, I used the particle under constant acceleration model (relevant equation I had posted) to find an equation for x as a function of time. From 0a, it is clear that the initial position x0 is 0 and that the initial velocity v0 is 0, leaving me with x = (1/2)axt2. From the slope, I found the acceleration from 0a to be about 3.33 m/s2 and upon plugging it into my equation, found x = 1.67t2.

Now to find an equation from ab, It is clear that the acceleration is 0 so it is convenient to use the particle under constant velocity model, x = x0 + vxt. I figured that since the final position from 0a is at t = 15s, I could plug in t = 15 to my first equation and obtain 375m and reason that the final position from 0a is the initial position from ab. It is also clear that the initial velocity from ab is 50 m/s, so I found that the equation from ab should be
x = 375 +50t. My text book says that the correct equation however, is x = -375 + 50t. Now I thought about this for a while, and I came to the conclusion that from 0a, 375m is indeed the distance traveled. However, the initial position is -375m because that is how FAR AWAY the car actually is from it's initial position... so x0 = -375 is just another way of expressing x0 = 0. (Not sure if my reasoning on that one is correct).

To find an equation for bc, I again figured I would use the particle under constant acceleration model. I calculated the slope of the line to find -5.00 m/s2, so after plugging it into the equation it is clear that the acceleration would be -2.5t2.. I figured the initial position would then be the total area under the curve from t = 0 to t = 40s which I found to be 1625m and using the same reasoning from the previous equation, I figured that the initial position would be X0 = -1625 m. It was also clear to me that the initial velocity v0 was 50 m/s again, leaving me with the equation,
x = 50t -2.5t2 - 1625. This also seems to be wrong and my book claims that the equation should be x = 250t -2.5t2 -4375.

I've been thinking about this problem for a while now but I can't quite seem to understand how these equations were formulated. I had just finished with a course in electricity and magnetism, but I am doing a lot of review to keep me on my toes. I would appreciate any help. This is also my first time posting, so if there is something wrong, please let me know and I will adjust accordingly. Thanks

Last edited by a moderator: May 6, 2017
2. Dec 21, 2013

Konoha

I must confess after reading your solution and following your steps, I was confused too, everything seemed right. Then it suddenly hit me! There is a tiny thing you didn't account for.

1. in part a, the total distance travelled isn't quiet right. (I reckon it might be a type) Check it again, and remember that the area of a Trapezoid is calculated by this formula: heightÃ—(base1+base2)/2

2. Part b and c are alright.

3. Now, about part d, the equation for 0-a was correct. Let's move to a-b, as you said the velocity is constant so we should use: xf = x0 + v0t, you calculated x0 from previous part, which is to be 375 m, and this is correct too. But here is the twist, imagine this, you are going to write an equation in which t=15 is actually t=0 for this part. so what you should do is to replace t in this new equation with t-15. (This way you make sure that whenever your input is 15s, it is actually t=0)
so we will have: xf = 375 + 50(t-15) --> xf= 50t-375

Apply this logic to next part and you should get the right answer for b-c.

Let me know how it goes. :)

3. Dec 22, 2013

Kyle_N22

Awesome, that was the missing link. Thank you for your help, it is much appreciated :).