Question

In an historical movie, two knights on horseback start from rest 92.3 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.245 m/s2, while Sir Alfred's has a magnitude of 0.227 m/s2. Relative to Sir George's starting point, where do the knights collide?

Answer 1

In an historical movie, two knights on horseback start from rest 92.3 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.245 m/s², while Sir Alfred's has a magnitude of 0.227 m/s². Relative to Sir George's starting point, where do the knights collide?

Solution

$a_{G}$ - Sir George's acceleration, $a_{A}$ - Sir Alfred's acceleration, t - time before collision,

$s$ - the distance between two knights, $s_G$ - the distance where do the knights collide relative to Sir George's starting point.

The space between the knights diminishes at acceleration

$$a = a _ {G} + a _ {A} = 0.245 \frac {\mathrm {m}}{\mathrm {s} ^ {2}} + 0.227 \frac {\mathrm {m}}{\mathrm {s} ^ {2}} = 0.472 \frac {\mathrm {m}}{\mathrm {s} ^ {2}}.$$

Collision occurs when $s = \frac{at^2}{2}$ :

$$s = \left(\frac {1}{2}\right) * 0.472 \frac {\mathrm {m}}{\mathrm {s} ^ {2}} * t ^ {2} = 92.3 \mathrm {m} \rightarrow t ^ {2} = 391.102 \rightarrow t = 19.8 \mathrm {s}.$$

George will have moved

$s_{G} = \frac{a_{G}t^{2}}{2} = \left(\frac{1}{2}\right)*0.245*19.8^{2} = 48m$ from his initial position at that time.

Answer: 47.908 m.