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 Post subject: Kinematics: 1DPosted: Sun Sep 22, 2013 6:57 pm

Joined: Fri Feb 20, 2009 6:38 pm
Silly, cocky me. I thought I had my 1D kinematics down from Grade 10. Come U1 Physics (remarkably boring, because I was a master at projectile motion), I decide to brush up. But apparently, I've forgotten a key point and I would like to check it out.

The situation is as follows:

Person 1 travels at 11.5 m/s and is 11.8 meters ahead of person 2. Person 2 travels at 9.7 m/s and accelerates at 1.2 m/s. I'm attempting to find the amount of time it takes for the second person to catch up

By my reasoning, the second person travels the same distance as the first person plus that 11.8m, giving d1=d and d2=d+11.8. They both take the same amount of time to cover their respective distances.

Is my reasoning correct? My solution depends on it...

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 Post subject: Re: Kinematics: 1DPosted: Sun Sep 22, 2013 8:40 pm

Joined: Thu Oct 04, 2012 8:49 pm
I believe it goes as such:
Person 1; x = v0 t + x0
Person 2; x = a t^2 + v0 t + x0
Set them equal to each other and solve for t.
x0 is initial position
v0 is initial velocity

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 Post subject: Re: Kinematics: 1DPosted: Mon Sep 23, 2013 12:52 pm

Joined: Sat Mar 26, 2011 9:56 am
quicksolver wrote:
By my reasoning, the second person travels the same distance as the first person plus that 11.8m, giving d1=d and d2=d+11.8. They both take the same amount of time to cover their respective distances.

Is my reasoning correct? My solution depends on it...

Your reasoning sounds correct to me.

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