Kinematics Physics (Relating velocity/time graph to displacement/time)

[__Shadow04__]

Homework Statement

The figure below represents part of the performance data of a car owned by a proud physics student. (The horizontal axis is marked in increments of 1 s and the vertical axis is marked in increments of 1 m/s.)

Here's a picture:
http://i.imgur.com/6bGXd.png

(d) Write an equation for x as a function of time for each phase of the motion, represented by the segments 0a, ab, and bc. (Use the following variable as necessary: t.)

Homework Equations

I know that I have to integrate the various line segments since the integration of velocity is displacement

The Attempt at a Solution

Now, treating the graph as a group of straight lines, in line segment 0a, I approximated the coordinates as (0,0) and (7.5,25) and got the slope, 3.33, since it's a line, the v(t) is 3.33t. Then, integrated in terms of t, and got (1/2)3.33t^2. That one was correct.

The problem comes when I try to get the displacement functions for ab and bc. Now, in line segment ab, I got that the function, since it's a straight line, v(t) is just 25. So, I thought the displacement function would just be 25t. But, apparently, I'm way off, since they got 25t+-93.75. Which...is confusing. I have no idea where they got that.

And in bc, I don't even have a clue what they did.

I think I'm missing something here. Anyone mind helping me out?

 

[LawrenceC]

They are determining the distance traveled for each phase of the three phases using absolute time.

 

[__Shadow04__]

And distance traveled is the area under the curve right?

So, wouldn't it be positive?

And if that's the case, what did they do to calculate x(t) for bc?

 

[LawrenceC]

For bc, if you plug in t=20, you will get t=406.25 which is also what you get if you plug in t=20 for the section ab. If you plug in t=7.5 for section ab, you get 93.75. If you plug in 7.5 in section ab, you get 93.75 if you use 3 and 1/3 rather than 3.33.

All three expressions provide positive displacements with continuity at times 7.5 and 20 seconds, respectively. Final distance is at t=25 seconds which is the area under the curve assuming the corner radii are made intersections of straight lines.

Time increments are 5 seconds, both axes.

 

[__Shadow04__]

Sorry for another question...

Okay, so I see that I have to use the 0a x(t) function I integrated in order to get the rest of these values.

I guess my only question left is, why do I have to plug in these numbers at all? Why isn't 25t for the interval [a,b] sufficient in determining the displacement?

Oh, and also, what formula did they use in the interval [b,c]? I know for [a,b] they used (I think), x=vt+x(initial) since it's under constant velocity on that interval.

But I don't seem to recognize the formula for [b,c]?

Thanks so much for the help. This is actually the last problem I need to solve.

 

[nDever]

Sorry for another question...

Okay, so I see that I have to use the 0a x(t) function I integrated in order to get the rest of these values.

I guess my only question left is, why do I have to plug in these numbers at all? Why isn't 25t for the interval [a,b] sufficient in determining the displacement?

Look at the v-t graph. From 0s to 7.5s, the object accelerated from rest to 25 m/s, so then, the average rate over that time interval was 12.5 m/s.

Think about the motion of an object starting at x=0. If from 0s to 7.5s the object was moving at an average rate of 12.5 m/s, where would it be at 7.5s? Insert 7.5 into the position function to get 93.75 m. In order to correctly represent the position at any time after 7.5s, you would have to start at 93.75 m because that's where the object was 7.5s after it accelerated from rest.

 

[__Shadow04__]

I get it. Haha, I guess I kinda answered my own question there by including x(initial).

I see.

But, what formula do they use for that last line segment? Is that the v^2=v(initial)^2+2a[x(final)-x(initial)]?

If not, how would I go about creating/finding that equation?

Thanks again.

 

[LawrenceC]

You would use

X = X0 + V0(t-20) + .5 * a * (t-20)^2

with

20 < t < 25

and a < 0

The less than signs for time should be less than or equal to. Can't type them on my computer.

X0 is initial displacement at t = 20. V0 is initial velocity which exists at the start of the interval 20 < t < 25. So V0 multiplied by the time after 20 seconds represents a distance.

The analogy is an object thrown into the air with initial velocity V0. The distance it goes up is

Y = Y0 + V0 * t + .5 *a * t^2 with 'a' being acceleration gravity < 0. Y0 is the thrower's initial elevation.