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August 8th, 2018, 05:17 AM   #1
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Motion Equations: Velocity Eliminate time?

I'm reading a Physics book and currently I'm in the Motion chapter. This chapter shows how you can calculate the motion equations. Below I'm going to give you a photo of what I did so far and inside a red rectangle you can see the equation which I can't calculate. The problem is to eliminate the time quantity for velocity's equation. I tried solving a second degree equation in order to find t and replace it into the velocity's equation, but I just could not solve it:


Last edited by skipjack; August 8th, 2018 at 09:55 AM.
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August 8th, 2018, 05:41 AM   #2
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Why not solve the first degree equation for $t$ instead?
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August 8th, 2018, 05:53 AM   #3
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Quote:
Originally Posted by Joppy View Post
Why not solve the first degree equation for $t$ instead?
LoL I'm stupid, forget about this thread

Why do I always think of the hard way first? And then getting so absorbed by it that I can't see there is a simpler way...

Last edited by skipjack; August 8th, 2018 at 09:55 AM.
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August 8th, 2018, 05:56 AM   #4
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By the way just from curiosity, is there a way to find time by solving a second degree equation and taking all the 3 possibilities from the distinctive (<0, =0, >0)? Or it's impossible? If it's possible, I would like to see a solution.

Last edited by skipjack; August 8th, 2018 at 09:56 AM.
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August 8th, 2018, 06:06 AM   #5
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Yes a good exercise. You may actually find more insight into what the variables describe
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August 8th, 2018, 07:01 AM   #6
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I did it!!! Can you take a look and answer my comments (inside the images)???

First Page:
https://image.ibb.co/bRe6Dz/first.jpg

Second Page:
https://image.ibb.co/iD3Ytz/second.jpg

Third Page:
https://image.ibb.co/jX07Le/third.jpg

Last edited by babaliaris; August 8th, 2018 at 07:04 AM.
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August 8th, 2018, 02:39 PM   #7
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$v_f = v_0 + at \implies t = \dfrac{v_f-v_0}{a}$


$\Delta x = v_0 \cdot t + \dfrac{1}{2}at^2$

substitute for $t$ ...

$\Delta x = v_0 \left(\dfrac{v_f-v_0}{a}\right) + \dfrac{a}{2} \left(\dfrac{v_f-v_0}{a}\right)^2$

$\Delta x = \dfrac{v_0v_f - v_0^2}{a} + \dfrac{a(v_f^2 - 2v_fv_0 + v_0^2)}{2a^2}$

$\Delta x = \dfrac{2v_0v_f - 2v_0^2}{2a} + \dfrac{v_f^2 - 2v_fv_0 + v_0^2}{2a}$

$\Delta x = \dfrac{2v_0v_f - 2v_0^2+v_f^2 - 2v_fv_0 + v_0^2}{2a}$

$2a \Delta x = v_f^2-v_0^2 \implies v_f^2 = v_0^2 + 2a \Delta x$
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