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I need more clearity on this physics problem

Posted on 2011-03-20
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I am new to this. I am trying to teach myself algebra and physics. I am trying to understand the last part of this under relating the objects. All circled variables are unknowns. More specifically part 3 and 4 especially 4, using a quadratic formula to figure out the unknown for t (time).

1. Can you provide some clarity how the author found the unknowns when there are multiple unknowns in each equation? I have read there are ways to solve for two unknowns, but in this case it's a little foggy to me.

2. When do I know a situation calls for the use of the quadratic equation? I had no idea to use that here.

t^2-18t-20=0

I would not have known what numbers and variables especially t^2 to insert into quadratic equation.

Thanks.
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Question by:kadin
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Expert Comment

by:ozo
ID: 35176951
1. one unknown was expressed in terms of the other using one equation, and then substituted into another equation

2. http://www.purplemath.com/modules/quadform.htm
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Author Comment

by:kadin
ID: 35177015
1. one unknown was expressed in terms of the other using one equation, and then substituted into another equation

Which equations are you referring to that this took place?
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by:ozo
ID: 35177083
XT expressed in terms of t from equation 2.2
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Author Comment

by:kadin
ID: 35177124
1. one unknown was expressed in terms of the other using one equation, and then substituted into another equation

XT expressed in terms of t from equation 2.2


Ok. That's the first equation. You said it was substituted into another equation. What's the other equation? Thanks.
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Expert Comment

by:ozo
ID: 35177156
XC = XT + 20
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by:TommySzalapski
ID: 35177799
When do I know a situation calls for the use of the quadratic equation?
If you have an equation that has a ^2 and you can't factor it, you need the quadratic equation. Technically, you could use it for any equation in the form ax^2 + bx + c = 0. For example, if you see x^2 = 9, you could say it's the same as 1x^2 + 0x - 9 = 0 and use the quadratic equation and you'll get the right answer, but it's a lot easier to use simpler methods.

t^2-18t-20=0

This doesn't factor nicely (if you try to factor it and can't, just use the quadratic).
So you have 1t^2 - 18t - 20 = 0
So use the quadratic equation with a = 1, b = -18 and c = -20
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Author Comment

by:kadin
ID: 35178599
Thanks for your input. That's good to know and it is helpful. I know how to factor and use the quadratic formula, but what I am really having a hard time understanding is how the author went from step 2 to 3 to 4.

I see in step 4 he created an equation that could be factored using the quadratic formula. It appears he brought elements down from the equation in step 2   - 18t - 20. Then he put this t^2 in front of it.

How or why did he know he should create this equation t^2-18t-20=0?

Where did he get t^2 from? Did he just pull it down from the equation in step 3 above? I don't think step 3 needs to be factored.

It is hard for me to follow the logic here. Thanks.
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Accepted Solution

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TommySzalapski earned 500 total points
ID: 35178640
It's one of the official starting equations. If you travel at a constant acceleration, your distance x will be 1/2*acceleration*time^2. But if you are already moving, then you need to add in the initial velocity*time. So:
x = v0t + 1/2at^2
So
Xc can be replaced with(Vc0t + 1/2Act^2)
Since the truck has constant velocity, you know the acceleration is 0 so XT = VTt.
This gives the equation in step 3.
Then you just plug in the values you know
(0)t + 1/2*(2)*t^2 = (18)t + 20
Then simple algebra gives step 4.
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Expert Comment

by:TommySzalapski
ID: 35178645
As for how he knew which of the equations to use: You just have to look carefully and be smart about it. Sometimes picking the wrong one will make the problem take five times as long even if it gets you the right answer.
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Author Comment

by:kadin
ID: 35183670
Then you just plug in the values you know
(0)t + 1/2*(2)*t^2 = (18)t + 20
Then simple algebra gives step 4.



How does simple algebra give step 4 t^2-18t-20=0?

If I subtract  -(18)t - 20

I get  (0)t + 1/2*(2)*t^2-(18)t - 20 =0

Not  t^2-18t-20=0

Thanks.
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Assisted Solution

by:TommySzalapski
TommySzalapski earned 500 total points
ID: 35200984
(0)t + 1/2*(2)*t^2-(18)t - 20 =0

(0)t = 0. Adding 0 does nothing so it's the same as
1/2*(2)*t^2-(18)t - 20 =0

1/2*2 = 1 so it's the same as
t^2-(18)t - 20 =0

Of course the () just imply multiplication so you have
t^2 - 18t - 20 = 0
Which is what you were trying to get.

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Author Closing Comment

by:kadin
ID: 35201345
Thanks for your ongoing help. You really help me learn this.
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