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Using logic, not formulas

  • Thread starter TrionHost
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  • #1
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Well, I am having trouble doing a big project that's due this monday. I can't get any of my formulas to work, and I so I used logic on a motion problem.
Here's my made up problem that I have to graph in Excel:
An object travels at 39m/s for 4 seconds and then slows down at a rate of -6m/s² for 6 seconds.
(Sorry the columns aren't that well aligned, I had to copy it from Excel)
Time, sec \ Position, m \ Velocity, m/s \ Acceleration, m/s2
0 ----------- 0 ----------- 0 ----------- 0
1 ----------- 39 ----------- 39 ----------- 0
2 ----------- 78 ----------- 39 ----------- 0
3 ----------- 117 ----------- 39 ----------- 0
4 ----------- 156 ----------- 39 ----------- 0
5 ----------- 189 ----------- 33 ----------- -6
6 ----------- 216 ----------- 27 ----------- -6
7 ----------- 237 ----------- 21 ----------- -6
8 ----------- 252 ----------- 15 ----------- -6
9 ----------- 261 ----------- 9 ----------- -6
10 ----------- 264 ----------- 3 ----------- -6
So..... I don't know any of the positions for sure after 4 seconds, or any of the velocities after 4 seconds. I just kind of used logic, it's pretty easy to follow what I did to find the positions and the velocities from 5-10 seconds. Are these right?
I can't get something that seems right when using my forumlas.... can you give me some forumals to find the position and the velocities?

(oh, and after reading this forums FAQ, Yes I have spent lots of hours on this project, and have gotten help from both my brother and dad, but I am not going to see them again before this project is due, so I really need some help)

THANKS A LOT!!!!!! :smile: :smile: :smile:
 
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Answers and Replies

  • #2
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What are you trying to graph? Velocity as a function of time? Displacement as a function of time?

I hate to tell you this too, but your chart is wrong. If we take the displacement as a function of time for the part of the problem where it is decelerating then we get [tex]s(t) = s_0 + v_{0}t + \frac{1}{2}at^2[/tex] where s_0 is the starting displacement (156m), v_0 is the starting velocity (39m/s), a is the acceleration (-6m/s^2), and t is the time.
[tex]\begin{multline*}s(1)&=156+39-3=192
\\ s(2)&=156+78-12=222
\\ s(3)&=156+117-27=246
\\ s(4)&=156+156-48=264
\\ s(5)&=156+195-75=276
\\ s(6)&=156+234-108=282
\end{multline*}[/tex]

I didn't use units in the interest of time, they cancel out to meters.

(Sorry, I'm still getting used to not doing typesetting in the LaTeX program. I tried using eqnarray, but didn't work out well, and I compromised with multiline, but if a moderator wants to fix it with the right command then they can.)
 
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  • #3
chroot
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I'd strongly suggest you give up on the "table" approach, and learn how to properly use the formula.

Can you show me the formulas you're using? What makes them appear to go wrong?

- Warren
 
  • #4
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I am graphing Velocity vs Time and Position vs Time

My equations: (from my brothers physics notecard, so they work apparently)

To find position: Xf=Xo+Vo+0.5at2 (that 2 is a squared)

To find velocity: V2=Vo2+2a(Xf-Xo) (the first two 2s are squared, last one is a actual 2)

Xf = final position
Xo = initial position
Vf = final velocity
Vo = initial velocity
t = time
a = accerleration

For solving for the 5 second mark, what is my inital position (0 or 156) and velocity (0 or 39)???
 
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  • #5
chroot
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That equation is actually NOT correct, TrionHost. The correct equation for the position at time t is:

[tex]x(t) = x_0 + v_0 t + \frac{1}{2} a t^2[/tex]

Note that you forgot the "t" on the [itex]v_0[/itex] term.

I'd also like to mention an easy way to verify that your formulas are correct: use the units. Note that an equation like [itex]x_0 = v_0[/itex] cannot possibly be correct, because position is measured in, for example, meters, while velocity is measured in meters per seconds. These two quantities cannot be equal, because their units are different, so this equation cannot be valid.

In the equation [itex]x(t) = x_0 + v_0 t + 1/2 a t^2[/itex], every term must have the same units. [itex]x_0[/itex] is obviously distance, as is velocity times time, as is acceleration times time squared.

- Warren
 
  • #6
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You have to divide it into two different problems. The first part is from the time 0s-4s where there is no acceleration and constant velocity. The second part is from the time 4s-10s where the acceleration is -6m/s^2 and a velocity decreasing linearly.
 
  • #7
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thanks chroot, learning a lot already....
about that equation, I just typed it wrong.

Now to Mindscrape...
So, when solving for the second half (when it accelerates) these are my numbers:
inital position 156 and velocity 39 - correct?
 
  • #8
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Correct.

The graph for displacement versus time should be a line with a positive slope until t=4, then from t=6 to t=10 gradually less and less of a slope. The velocity is the slope of the displacement at very small intervals of time, so it should have a straight y=39 line from t=0 to t=4 and a negative slope of y=-6x from t=6 to t=10.

It's really helpful to take a quick crash course in calculus. The site has good introductory lessons too.
 
  • #9
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about that calculus.... lol, I'm in 9th grade. Thanks guys.
 
  • #10
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when doing this to find the velocity:
Xf=Xo+Vo+0.5at2 (that 2 is a squared)

the equations turns out like this:
156+39*5+0.5*-6*25=276

how can the position go from 156 to 276 (120 difference), when the inital velocity was 39, and it was decelerating? Did I do something wrong?
 
  • #11
chroot
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That sounds correct to me. If the object were not decelerating, it would have gone 156 units instead of 120 units.

- Warren
 
  • #12
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well ok, but then I worked out the initial velocity for when it begins to decelerate, and I got 9.

39*39+2*-6(276-156)=9

Does that make since for it to drop from 39 to 9?

and obiviously when I keep going I am going to get a negative velocity, does anything special happen then or does the position just go in reverse?

thanks again.
 
  • #13
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well I figured out the the velocity does in fact go to 9.

a=Vf-Vo/Tf-To - just would the Initial Time be 0 or 4? 0 is what makes it work, but if I am working for the part when it decelerates shouldn't it be 4 seconds?

o, and again, what happens when I get to a negative velocity?
 
  • #14
chroot
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You had already found the initial velocity before it begins to decelerate; it was given in the problem: 39 m/s.

- Warren
 
  • #15
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i think you misunderstood me.
 
  • #16
chroot
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Feel free to explain in more detail, then. I thought the problem was completed.

- Warren
 
  • #17
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here's my info so far:
Time, sec Position, m Velocity, m/s Acceleration, m/s2
0 --------- 0 --------- 0 --------- 0
1 --------- 39 --------- 39 --------- 0
2 --------- 78 --------- 39 --------- 0
3 --------- 117 --------- 39 ---------0
4 --------- 156 --------- 39 --------- 0
*5 --------- 276 --------- 9 --------- -6
6 --------- 282 --------- 9 --------- -6
7 --------- 282 --------- 9 --------- -6
8 --------- 276 --------- ? --------- -6
9 --------- 264 --------- ? --------- -6
10 --------- 246 --------- ? --------- -6

for the position and velocity for 5 seconds on down are what I am having trouble with. How can 6 and 7 seconds have the same position? I check my forumla, and they turn out the same, and how come the velocity isn't changing? I must be doing something wrong.

I am using those same equations as mentioned earlier and
Vf=Vo+at - same thing happens
Vf=atf+(Vo-ato) - "

Please give me some feedback.
 
  • #18
chroot
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Why on earth do you think the velocity is discontinuous, going from 39 m/s to 9 m/s in a single second?

The deceleration is given as -6 m/s/s, meaning that the velocity decreases by 6 m/s every second. It's 39 m/s at t=4, and 39-6=33 m/s at t=5, and 33-6=27 m/s at t=6, etc.

You really need to get away from these tables and look at the equations!

- Warren
 
  • #19
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So these would be my Velocities (starting at 4 seconds):
0
39
39
39
39
33
27
21
15
9
3

So I don't need to use a forumula for the velocity in this case? and now should I use a forumla too for the positions, or can I just keep adding the velocity to the last position?
 
  • #20
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Well, essentially you are using a formula. The final velocity is the initial velocity plus the negative acceleration as a factor of time.

[tex]v(t)=v_i - at[/tex]

So after one second the velocity decreases by six from the initial, and after two seconds the velocity decreases by twelve from the initial, etc.
 
  • #21
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I think I found my problem!!!

When it decelerates, I have to reset the time back to 0 in the equations, instead of 4 seconds. All this time I used 4 for inital time instead of 0, because the 4 second mark is the begining of the deceleration.
 
  • #22
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Do my graphs look right, I think it's finished:

http://www.trionhost.com/graphs.JPG [Broken]

THANKS SO MUCH!!!
I am definetly going to stick around these forums...
 
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  • #23
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TrionHost said:
I think I found my problem!!!

When it decelerates, I have to reset the time back to 0 in the equations, instead of 4 seconds. All this time I used 4 for inital time instead of 0, because the 4 second mark is the begining of the deceleration.
Yes, they have to different circumstances so you have to split the time intervals into two. If you have seen piecewise functions before then that is what this is.

Your graphs are right, but the velocity versus time would be better if you could connect the two, since it is continuous.
 

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