Question #239233

In a historical movie, two knights on horseback start from rest 88.0 m apart and ride directly toward each other to do battle. Sir George’s acceleration has a magnitude of 0.300 m/s2 , while Sir Alfred’s has a magnitude of 0.200 m/s2 . Relative to Sir George’s starting point, where do the knights collide?



1
Expert's answer
2021-09-20T10:00:46-0400

Let L=88 mL=88\ m is the distance between two knights before the start, dd is the distance from Sir George’s starting point to the point where the knights collide. Then, we can write the displacements of each knights:


dG=12aGt2=d,d_G=\dfrac{1}{2}a_Gt^2=d,dA=12aAt2=Ld.d_A=\dfrac{1}{2}a_At^2=L-d.

Let's find t2t^2 from the first equation and plug it into the second one:


t2=2daG,t^2=\dfrac{2d}{a_G},12aA(2daG)=Ld,\dfrac{1}{2}a_A(\dfrac{2d}{a_G})=L-d,d(aAaG)=Ld.d(\dfrac{a_A}{a_G})=L-d.

Solving for dd we get:


d(aAaG+1)=L,d(\dfrac{a_A}{a_G}+1)=L,d=LaAaG+1,d=\dfrac{L}{\dfrac{a_A}{a_G}+1},d=88 m0.2 ms20.3 ms2+1=52.8 m.d=\dfrac{88\ m}{\dfrac{0.2\ \dfrac{m}{s^2}}{0.3\ \dfrac{m}{s^2}}+1}=52.8\ m.

Answer:

d=52.8 m.d=52.8\ m.

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