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March 18th, 2019, 03:14 PM   #1
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A simple Proof by the Easy Or Hard Way?

Given the equations below prove that $\displaystyle v^2 = v_0^2 + 2a(x-x_0)$

1) $\displaystyle
v = v_0 + at
$
2) $\displaystyle
x-x_0 = v_ot + \frac{1}{2}at^2
$


Easy Solution:
$\displaystyle
v = v_0 + at \Leftrightarrow t = \frac{v-v_0}{a}
$

Substituting t into equation 2
$\displaystyle
\Delta{x} = v_0\frac{v-v_0}{a} + \frac{1}{2}a\frac{(v-v_0)^2}{a^2}
= \frac{vv_0 - v_0^2}{a} + \frac{1}{2}\frac{v^2 - 2vv_0 + v_0^2}{a}
$

Multiplying both sides with 2a

$\displaystyle
2a\Delta{x} = 2vv_0 - 2v_0^2 + v^2 - 2vv_0 + v_0^2 = v^2 - v_0^2 \Leftrightarrow
v^2 = v_0^2 + 2a\Delta{x} \Leftrightarrow v^2 = v_0^2 + 2a(x-x_0)
$



Difficult Solution:
$\displaystyle
\Delta{x} = v_ot + \frac{1}{2}at^2 \Leftrightarrow \frac{1}{2}at^2 + v_ot - \Delta{x} = 0
$

$\displaystyle
d = v_0^2 - 4\frac{1}{2}a\cdot(-\Delta{x}) \Leftrightarrow d = v_0^2 + 2a\Delta{x}
$

Case 1: d < 0

$\displaystyle
t_{1,2} = \frac{-v_0 \pm j\sqrt{|v_0^2 + 2a\Delta{x}|} }{2\frac{1}{2}a} =
\frac{-v_0 \pm j\sqrt{|v_0^2 + 2a\Delta{x}|} }{a}
$

Case 2: d = 0

$\displaystyle
t = \frac{-v_0} {2\frac{1}{2}a} =
\frac{-v_0}{a}
$


Case 3: d > 0

$\displaystyle
t_{1,2} = \frac{-v_0 \pm \sqrt{v_0^2 + 2a\Delta{x}} }{2\frac{1}{2}a} =
\frac{-v_0 \pm \sqrt{v_0^2 + 2a\Delta{x}} }{a}
$


Because $\displaystyle v_0^2 > 0$ and $\displaystyle \Delta{x} > 0$ (Can't have negative length in physics)
$\displaystyle
d > 0 \Leftrightarrow v_0^2 + 2a\Delta{x} > 0 \Leftrightarrow a > 0
$ THEN

$\displaystyle
t_{2} = \frac{-v_0 - \sqrt{v_0^2 + 2a\Delta{x}} }{a} < 0
$ is rejected as a solution because we can't have a negative time.


Substituting $\displaystyle t_{1} = \frac{-v_0 + \sqrt{v_0^2 + 2a\Delta{x}} }{a}$ into equation 1

$\displaystyle
v = v_0 + at = v_0 + a\frac{-v_0 + \sqrt{v_0^2 + 2a\Delta{x}} }{a} =
v_0 - v_0 + \sqrt{v_0^2 + 2a\Delta{x}} = \sqrt{v_0^2 + 2a\Delta{x}}
$

Squaring both sides
$\displaystyle
v^2 = \sqrt{v_0^2 + 2a\Delta{x}}^2 = |v_0^2 + 2a\Delta{x}| \Leftrightarrow
v^2 = v_0^2 + 2a(x-x_0)
$


First Question
The problem is that I don't know what to say about the other solutions in case 1 and 2.
Do I reject them for some reason? I can't see the answer to that.


Second Question
The answer is $\displaystyle v^2 = v_0^2 + 2a(x-x_0)$ and v is a one dimensional vector (Velocity). Because the velocity is squared in the above equation does that mean that this form can only tell as the magnitude of the velocity and not the direction?

Last edited by babaliaris; March 18th, 2019 at 03:26 PM.
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March 18th, 2019, 03:21 PM   #2
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The case 3 gives the equation but a can be negative or positive .
a would be negative if $\displaystyle v_0 > v$ .

You have the direction in the basic equation $\displaystyle v=v_0 +at$ .
Thanks from topsquark

Last edited by idontknow; March 18th, 2019 at 03:32 PM.
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March 18th, 2019, 03:27 PM   #3
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Quote:
Originally Posted by babaliaris View Post
Given the equations below prove that $\displaystyle v^2 = v_0^2 + 2a(x-x_0)$

1) $\displaystyle
v = v_0 + at
$
2) $\displaystyle
x-x_0 = v_ot + \frac{1}{2}at^2
$


Easy Solution:
$\displaystyle
v = v_0 + at \Leftrightarrow t = \frac{v-v_0}{a}
$

Substituting t into equation 2
$\displaystyle
\Delta{x} = v_0\frac{v-v_0}{a} + \frac{1}{2}a\frac{(v-v_0)^2}{a^2}
= \frac{vv_0 - v_0^2}{a} + \frac{1}{2}\frac{v^2 - 2vv_0 + v_0^2}{a}
$

Multiplying both sides with 2a

$\displaystyle
2a\Delta{x} = 2vv_0 - 2v_0^2 + v^2 - 2vv_0 + v_0^2 = v^2 - v_0^2 \Leftrightarrow
v^2 = v_0^2 + 2a\Delta{x} \Leftrightarrow v^2 = v_0^2 + 2a(x-x_0)
$



Difficult Solution:
$\displaystyle
\Delta{x} = v_ot + \frac{1}{2}at^2 \Leftrightarrow \frac{1}{2}at^2 + v_ot - \Delta{x} = 0
$

$\displaystyle
d = v_0^2 - 4\frac{1}{2}a\cdot(-\Delta{x}) \Leftrightarrow d = v_0^2 + 2a\Delta{x}
$

Case 1: d < 0

$\displaystyle
t_{1,2} = \frac{-v_0 \pm j\sqrt{|v_0^2 + 2a\Delta{x}|} }{2\frac{1}{2}a} =
\frac{-v_0 \pm j\sqrt{|v_0^2 + 2a\Delta{x}|} }{a}
$

Case 2: d = 0

$\displaystyle
t = \frac{-v_0} {2\frac{1}{2}a} =
\frac{-v_0}{a}
$


Case 3: d > 0

$\displaystyle
t_{1,2} = \frac{-v_0 \pm \sqrt{v_0^2 + 2a\Delta{x}} }{2\frac{1}{2}a} =
\frac{-v_0 \pm \sqrt{v_0^2 + 2a\Delta{x}} }{a}
$


Because $\displaystyle v_0^2 > 0$ and $\displaystyle \Delta{x} > 0$ (Can't have negative length in physics)
$\displaystyle
d > 0 \Leftrightarrow v_0^2 + 2a\Delta{x} > 0 \Leftrightarrow a > 0
$ THEN

$\displaystyle
t_{2} = \frac{-v_0 - \sqrt{v_0^2 + 2a\Delta{x}} }{a} < 0
$ is rejected as a solution because we can't have a negative time.


Substituting $\displaystyle t_{1} = \frac{-v_0 + \sqrt{v_0^2 + 2a\Delta{x}} }{a}$ into equation 1

$\displaystyle
v = v_0 + at = v_0 + a\frac{-v_0 + \sqrt{v_0^2 + 2a\Delta{x}} }{a} =
v_0 - v_0 + \sqrt{v_0^2 + 2a\Delta{x}} = \sqrt{v_0^2 + 2a\Delta{x}}
$

Squaring both sides
$\displaystyle
v^2 = \sqrt{v_0^2 + 2a\Delta{x}}^2 = |v_0^2 + 2a\Delta{x}| \Leftrightarrow
v^2 = v_0^2 + 2a(x-x_0)
$



The problem is that I don't know what to say about the other solutions in case 1 and 2.
Do I reject them for some reason? I can't see the answer to that.
d < 0. The times are complex numbers!

d = 0. This implies that v = 0, which is not true in general.

-Dan

PS Nice work, by the way.
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March 18th, 2019, 03:30 PM   #4
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Quote:
Originally Posted by topsquark View Post
d < 0. The times are complex numbers!

d = 0. This implies that v = 0, which is not true in general.

-Dan

PS Nice work, by the way.
So times can not be complex numbers?
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March 18th, 2019, 03:33 PM   #5
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Quote:
Originally Posted by babaliaris View Post
The answer is $\displaystyle v^2 = v_0^2 + 2a(x-x_0)$ and v is a one dimensional vector (Velocity). Because the velocity is squared in the above equation does that mean that this form can only tell as the magnitude of the velocity and not the direction?
Sorry, I missed this one.

Unless we are working with something more advanced (like Relativity) we can assume that when we see $\displaystyle v^2$ then we actually mean $\displaystyle v^2 = \vec{v} \cdot \vec{v}$, where $\displaystyle \cdot$ is the usual dot product. (This notation is true for any vector.)

Notice that $\displaystyle v^2 = \vec{v} \cdot \vec{v}$ with no angles involved for any number of dimensions as the angle between any vector and itself is always 0.

-Dan
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March 18th, 2019, 03:34 PM   #6
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Quote:
Originally Posted by babaliaris View Post
So times can not be complex numbers?
We're crossing posts!

No, times are never complex numbers. Times are always taken to be real numbers. If you get a complex number for an answer to just about anything in Introductory Physics then you've done something wrong. In this case we can't have d < 0.

-Dan
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March 18th, 2019, 03:50 PM   #7
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Awesome answers thank you! I learned something new today.
I'm an engineer student but I love to use math strictly, even that in Engineering we don't usually do that.
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