Solve for Knights' Collision Point

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If someone could help me out with this problem I would greatly appreciate it.

Thanks in advance.

In an historical movie, two knights on horseback start from rest 81.5 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.207 m/s2, while Sir Alfred's has a magnitude of 0.339 m/s2. Relative to Sir George's starting point, where do the knights collide?
 
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Hi skud363,
Welcome to PF! We're here to help you out...show what you've done so far, or your ideas on how to start solving.
 
I understand that the formula x=1/2 (vi+vf)t^2 has to be applied. I am able to find the final velocities easily, but the problem is if they are colliding at one certain point, that means that both of their displacements has to be equal. And since I'm not sure of how much time has elapsed since they are both beginning at rest, it's just a matter of rearranging the formula but I'm not sure how to do that. Please reply w/ any suggestions/detailed explanations, etc...
 
Well I have a neat way of thinking of it. I am not 100% sure that it will work, but I am 99.9%. (I am in a grade 11 physics course). Your horses have to run a total of 81.5m altogether to meet with one another. So think of it like this, if you have both horses start at the exact same point, how would you find out how long until their TOTAL distance (horse A distance + horse B distance) = 81.5m

I think you have to work through a few equations to get your answer.

"but the problem is if they are colliding at one certain point, that means that both of their displacements has to be equal" - This is not true, if one is moving faster than the other, it will travel farther than the other before they colide.

Also, use ^ to indicate a power. 0.207m/s^2 etc ;)
 
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They are not accelerating at the same rate and therefore they wouldn't have traveled the same distance when they meet.

Here's what you do,
Let Sa be the distance that Alfred has traveled and Sg for George. You know their initial velocities. Now, apply the general formula s = ut + 1/2at^2 for both knights. (How long does each knight take to reach the point where the collide?)