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in an historical movie, two knights on horseback start from rest 95.3 m apart and ride directly toward each other to do battle. sir georges acceleration has a magnitude of 0.296 m/s², while sir alfreds has a magnitude of 0.315 m/s². relative to sir georges starting point, where do the knights collide?
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Explanation:

Step1: Set up equations for displacement

Let the distance Sir George travels be $x_G$ and the distance Sir Alfred travels be $x_A$. The initial distance between them is $d = 95.3$ m, so $x_G+x_A=d$. Using the equation of motion $x = v_0t+\frac{1}{2}at^{2}$, since they start from rest ($v_0 = 0$), $x_G=\frac{1}{2}a_Gt^{2}$ and $x_A=\frac{1}{2}a_At^{2}$, where $a_G = 0.296$ m/s² and $a_A=0.315$ m/s².

Step2: Express the relationship in terms of acceleration

Substitute $x_G$ and $x_A$ into $x_G + x_A=d$: $\frac{1}{2}a_Gt^{2}+\frac{1}{2}a_At^{2}=d$. Factor out $\frac{1}{2}t^{2}$ to get $\frac{1}{2}t^{2}(a_G + a_A)=d$. Then solve for $t^{2}$: $t^{2}=\frac{2d}{a_G + a_A}$.

Step3: Calculate the time $t$

Substitute $d = 95.3$ m, $a_G = 0.296$ m/s² and $a_A=0.315$ m/s² into the formula for $t^{2}$: $t^{2}=\frac{2\times95.3}{0.296 + 0.315}=\frac{190.6}{0.611}\approx312$. Then $t=\sqrt{312}\approx17.66$ s.

Step4: Calculate the position of collision relative to Sir George

Use the equation $x_G=\frac{1}{2}a_Gt^{2}$. Substitute $a_G = 0.296$ m/s² and $t\approx17.66$ s: $x_G=\frac{1}{2}\times0.296\times(17.66)^{2}=\frac{1}{2}\times0.296\times312\approx46.3$ m.

Answer:

46.3 m