How Do Two Knights Collide When Accelerating Towards Each Other?

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Two knights, Sir George and Sir Alfred, start 88.0 meters apart and accelerate towards each other with different rates: 0.300 m/s² and 0.200 m/s², respectively. The discussion involves calculating the point of collision relative to Sir George's starting position using kinematic equations. The key formula for displacement in terms of time and acceleration is highlighted, leading to the conclusion that the knights will collide at a specific distance from Sir George's starting point. The calculations focus on determining the time of collision based on their accelerations. Ultimately, the problem is resolved through the application of these kinematic principles.
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1. two knights on horseback start from rest 88.0 m apart and ride directly toward each other to do battle. Sir George's acceleation has a magnitude of 0.300 m/s^2, Sir Alfred's has a magnitude of 0.200 m/s^2. Relative to Sir George's starting point, where do the knights collide?



2. a= (v-v0)/(t-t0), v=(x-x0)/(t-t0)



3. 0.300m/s^2=(v-0)/t, v=0.300m/s^2*t, 0.300m/s^2*t=(x-x0)/t, I don't know where to go from here!
 
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You have a formula for displacement in terms of time and acceleration... try to use that to get the time when they collide...
 
x=(0.300m/s^2 *t^2)/2 ?
 
nevermind, I got it now!
 

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