Answer:
Relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.
Explanation:
Let the distance covered by Sir George be
and the distance covered by Sir Alfred be 
Since the knights collide, hence they must have traveled for the same amount of time just before collision
From one of the equations of motion for linear motion

Where
is the distance traveled
is the initial velocity
is the acceleration
and
is the time
For Sir George,

= 0 m/s (Since they start from rest)
0.21 m/s²
Hence,
becomes


Now, for Sir Alfred

= 0 m/s (Since they start from rest)
0.26 m/s²
Hence,
becomes


Since, they traveled for the same time,
just before collision, we can write

Since, the two nights are 86 m apart, that is, the sum of the distances covered by the knights just before collision is 86 m. Then we can write that
m
Then, 
Then,
becomes


m
∴ Sir George covered a distance of 38.43 m just before collision.
Hence, relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.