Answer:
The knights will collide at 43.854 meters relative to Sir George's starting point.
Explanation:
Let suppose that initial positions of Sir George and Sir Alfred are 0 and 84.1 meters, respectively. If both knights accelerate uniformly, then we have the following kinematic formulas:
Sir George
[tex]x_{G} = x_{G,o}+v_{o,G}\cdot t + \frac{1}{2}\cdot a_{G}\cdot t^{2}[/tex] (1)
Sir Alfred
[tex]x_{A} = x_{A,o}+v_{o,A}\cdot t + \frac{1}{2}\cdot a_{A}\cdot t^{2}[/tex] (2)
Where:
[tex]x_{G,o}[/tex], [tex]x_{A,o }[/tex] - Initial position of Sir George and Sir Alfred, measured in meters.
[tex]x_{G}[/tex], [tex]x_{A}[/tex] - Final position of Sir George and Sir Alfred, measured in meters.
[tex]v_{o,G}[/tex], [tex]v_{o,A}[/tex] - Initial velocity of Sir George and Sir Alfred, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]a_{G}[/tex], [tex]a_{A}[/tex] - Acceleration of Sir George and Sir Alfred, measured in meters per square second.
Both knights collide when [tex]x_{G} = x_{A}[/tex], then we simplify this system of equations below:
[tex]x_{G,o} + v_{o,G}\cdot t + \frac{1}{2}\cdot a_{G}\cdot t^{2} = x_{A,o}+v_{o,A}\cdot t + \frac{1}{2}\cdot a_{A}\cdot t^{2}[/tex]
[tex](x_{A,o}-x_{G,o}) +(v_{o,A}-v_{o,G})\cdot t +\frac{1}{2}\cdot (a_{A}-a_{G})\cdot t^{2} = 0[/tex] (3)
If we know that [tex]x_{A,o} = 84.1\,m[/tex], [tex]x_{G,o} = 0\,m[/tex], [tex]v_{o,A} = 0\,\frac{m}{s}[/tex], [tex]v_{o,G} = 0\,\frac{m}{s}[/tex], [tex]a_{A} = -0.289\,\frac{m}{s^{2}}[/tex] and [tex]a_{G} = 0.316\,\frac{m}{s^{2}}[/tex], then we have the following formula:
[tex]84.1 -0.303\cdot t^{2} = 0[/tex] (4)
The time associated with collision is:
[tex]t \approx 16.660\,s[/tex]
And the point of collision is:
[tex]x_{G} = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (16.660\,s)+ \frac{1}{2}\cdot \left(0.316\,\frac{m}{s^{2}} \right) \cdot (16.660\,s)^{2}[/tex]
[tex]x_{G} = 43.854\,m[/tex]
The knights will collide at 43.854 meters relative to Sir George's starting point.
