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 March 18th, 2019, 04:14 PM #1 Senior Member   Joined: Oct 2015 From: Greece Posts: 139 Thanks: 8 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 04:26 PM. March 18th, 2019, 04:21 PM #2 Senior Member   Joined: Dec 2015 From: Earth Posts: 820 Thanks: 113 Math Focus: Elementary Math 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 04:32 PM. March 18th, 2019, 04:27 PM   #3
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Quote:
 Originally Posted by babaliaris 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.  March 18th, 2019, 04:30 PM   #4
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Quote:
 Originally Posted by topsquark 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? March 18th, 2019, 04:33 PM   #5
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 Originally Posted by babaliaris 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 March 18th, 2019, 04:34 PM   #6
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 Originally Posted by babaliaris 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 March 18th, 2019, 04:50 PM #7 Senior Member   Joined: Oct 2015 From: Greece Posts: 139 Thanks: 8 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. Tags easy, hard, proof, simple Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post flei Algebra 3 December 3rd, 2014 08:04 AM maple_leaf Elementary Math 9 May 25th, 2013 07:41 AM nukem4111 Number Theory 1 September 18th, 2012 11:52 AM lenfromkits Calculus 4 November 23rd, 2011 11:27 AM foo_ Computer Science 1 April 20th, 2009 11:47 AM

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