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One dimensional kinematics problem...?
In a historical movie, two knights on horseback start from rest 95 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.18 m/s2, while Sir Alfred's has a magnitude of 0.26 m/s2. Relative to Sir George's starting point, where do the knights collide?
If you can solve it, can you show the equations that you need to use and how you got there?
Thanks!!
1 Answer
- DHLv 71 decade agoFavorite Answer
Let x = distance Sir George travels to the impacr..then (95 - x) equals Sir Alfred's distance
Now starting from rest the distance each covers is x = 1/2*a*t^2
Since they each take the same time to reach the impact point we have
S.G. x = 1/2*0.18*t^2 or t^2 = 2*x/0.18
S.A. (95 -x) = 1/2*0.26*t^2 or t^2 = 2*(95-x)/0.26
Now we equate these giving 2*x/0.18 = 2*(95-x)/0.26
Simplifying we get 11.11x = 731 - 7.69x...so (11.11 + 7.69)*x = 731
So x = 731/(11.11+7.69) = 38.9m