Respuesta :
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 [tex]S_{1}[/tex]
and the distance covered by Sir Alfred be [tex]S_{2}[/tex]
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
[tex]S = ut + \frac{1}{2}at^{2}[/tex]
Where [tex]S[/tex] is the distance traveled
[tex]u[/tex] is the initial velocity
[tex]a[/tex] is the acceleration
and [tex]t[/tex] is the time
For Sir George,
[tex]S = S_{1}[/tex]
[tex]u[/tex] = 0 m/s (Since they start from rest)
[tex]a =[/tex]0.21 m/s²
Hence,
[tex]S = ut + \frac{1}{2}at^{2}[/tex] becomes
[tex]S_{1} = (0)t + \frac{1}{2}(0.21)t^{2}\\S_{1} = 0.105 t^{2}\\[/tex]
[tex]t^{2} = \frac{S_{1}}{0.105}[/tex]
Now, for Sir Alfred
[tex]S = S_{2}[/tex]
[tex]u[/tex] = 0 m/s (Since they start from rest)
[tex]a =[/tex]0.26 m/s²
Hence,
[tex]S = ut + \frac{1}{2}at^{2}[/tex] becomes
[tex]S_{2} = (0)t + \frac{1}{2}(0.26)t^{2}\\S_{2} = 0.13 t^{2}\\[/tex]
[tex]t^{2} = \frac{S_{2}}{0.13}[/tex]
Since, they traveled for the same time, [tex]t[/tex] just before collision, we can write
[tex]\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}[/tex]
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
[tex]S_{1} + S_{2} = 86[/tex] m
Then, [tex]S_{2} = 86 - S_{1}[/tex]
Then,
[tex]\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}[/tex] becomes
[tex]\frac{S_{1}}{0.105}= \frac{86 -S_{1}}{0.13}[/tex]
[tex]0.13{S_{1}}= 0.105({86 -S_{1}})\\0.13{S_{1}}= 9.03 - 0.105S_{1}}\\0.13{S_{1}} + 0.105S_{1}}= 9.03 \\0.235{S_{1}} = 9.03\\{S_{1}} =\frac{9.03}{0.235}[/tex]
[tex]S_{1} = 38.43[/tex] 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.
