- #1

Kyle_N22

- 2

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## Homework Statement

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

## Homework Equations

x

_{f}= x

_{0}+ v

_{0}t + (1/2)a

_{x}t

^{2}

x

_{f}= x

_{0}+ v

_{0}t

## 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 x

_{0}is 0 and that the initial velocity v

_{0}is 0, leaving me with x = (1/2)a

_{x}t

^{2}. From the slope, I found the acceleration from 0a to be about 3.33 m/s

^{2}and upon plugging it into my equation, found x = 1.67t

^{2}.

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 = x

_{0}+ v

_{x}t. 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 textbook 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 x

_{0}= -375 is just another way of expressing x

_{0}= 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/s

^{2}, so after plugging it into the equation it is clear that the acceleration would be -2.5t

^{2}.. 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 X

_{0}= -1625 m. It was also clear to me that the initial velocity v

_{0}was 50 m/s again, leaving me with the equation,

x = 50t -2.5t

^{2}- 1625. This also seems to be wrong and my book claims that the equation should be x = 250t -2.5t

^{2}-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

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