My Math Forum Motion Equations: Velocity Eliminate time?

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 August 8th, 2018, 05:17 AM #1 Member   Joined: Oct 2015 From: Greece Posts: 81 Thanks: 6 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.
 August 8th, 2018, 05:41 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,765 Thanks: 623 Math Focus: Yet to find out. Why not solve the first degree equation for $t$ instead?
August 8th, 2018, 05:53 AM   #3
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Quote:
 Originally Posted by Joppy Why not solve the first degree equation for $t$ instead?

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.

 August 8th, 2018, 05:56 AM #4 Member   Joined: Oct 2015 From: Greece Posts: 81 Thanks: 6 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.
 August 8th, 2018, 06:06 AM #5 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,765 Thanks: 623 Math Focus: Yet to find out. Yes a good exercise. You may actually find more insight into what the variables describe Thanks from babaliaris
 August 8th, 2018, 07:01 AM #6 Member   Joined: Oct 2015 From: Greece Posts: 81 Thanks: 6 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.
 August 8th, 2018, 02:39 PM #7 Math Team   Joined: Jul 2011 From: Texas Posts: 2,818 Thanks: 1463 $v_f = v_0 + at \implies t = \dfrac{v_f-v_0}{a}$ $\Delta x = v_0 \cdot t + \dfrac{1}{2}at^2$ substitute for $t$ ... $\Delta x = v_0 \left(\dfrac{v_f-v_0}{a}\right) + \dfrac{a}{2} \left(\dfrac{v_f-v_0}{a}\right)^2$ $\Delta x = \dfrac{v_0v_f - v_0^2}{a} + \dfrac{a(v_f^2 - 2v_fv_0 + v_0^2)}{2a^2}$ $\Delta x = \dfrac{2v_0v_f - 2v_0^2}{2a} + \dfrac{v_f^2 - 2v_fv_0 + v_0^2}{2a}$ $\Delta x = \dfrac{2v_0v_f - 2v_0^2+v_f^2 - 2v_fv_0 + v_0^2}{2a}$ $2a \Delta x = v_f^2-v_0^2 \implies v_f^2 = v_0^2 + 2a \Delta x$ Thanks from babaliaris

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