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March 18th, 2019, 03:14 PM  #1 
Senior Member Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8  A simple Proof by the Easy Or Hard Way?
Given the equations below prove that $\displaystyle v^2 = v_0^2 + 2a(xx_0)$ 1) $\displaystyle v = v_0 + at $ 2) $\displaystyle xx_0 = v_ot + \frac{1}{2}at^2 $ Easy Solution: $\displaystyle v = v_0 + at \Leftrightarrow t = \frac{vv_0}{a} $ Substituting t into equation 2 $\displaystyle \Delta{x} = v_0\frac{vv_0}{a} + \frac{1}{2}a\frac{(vv_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(xx_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(xx_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 $\displaystyle v^2 = v_0^2 + 2a(xx_0)$ 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. 
March 18th, 2019, 03:21 PM  #2 
Senior Member Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 
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 $\displaystyle v=v_0 +at$ . Last edited by idontknow; March 18th, 2019 at 03:32 PM. 
March 18th, 2019, 03:27 PM  #3  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,266 Thanks: 934 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
d = 0. This implies that v = 0, which is not true in general. Dan PS Nice work, by the way.  
March 18th, 2019, 03:30 PM  #4 
Senior Member Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8  
March 18th, 2019, 03:33 PM  #5  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,266 Thanks: 934 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
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, 03:34 PM  #6 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,266 Thanks: 934 Math Focus: Wibbly wobbly timeywimey stuff.  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, 03:50 PM  #7 
Senior Member Joined: Oct 2015 From: Greece Posts: 137 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. 

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