Question

In a historical movie, two knight on horseback start from

rest88.0 m apart and ride directly toward each other to dobattle.
Sir George's acceleration has a magnitude of0.300 m/s2
while Sir Alfred's has a magnitude of0.200 m/s2
Relative to Sir George's starting point, where do the
knightscollide?

Answer 1

The knights collide 53.0 m from the starting point of sir George.

Step-by-step explanation:

The equation for the position in a straight accelerated movement is as follows:

x = x0 + v0 t + 1/2 a t²

where

x = position at time t

x0 = initial position

v0 = initial speed

a = acceleration

t = time

The position of the two knights is the same when they collide. Since they start from rest, v0 = 0:

Sir George´s position:

xGeorge = 0 m + 0 m + 1/2 * 0.300 m/s² * t²

Considering the center of the reference system as Sir George´s initial position, the initial position of sir Alfred will be 88.0 m. The acceleration of sir Afred will be negative because he rides in opposite direction to sir George:

xAlfred = 88.0 m + 0 m - 1/2 * 0.200 m/s² * t²

When the knights collide:

xGeorge = x Alfred

1/2 * 0.300 m/s² * t² = 88.0 m - 1/2 * 0.200 m/s² * t²

0.150 m/s² * t² = 88.0 m - 0.100 m/s² * t²

0.150 m/s² * t² + 0.100 m/s² * t² = 88.0 m

0.250 m/s² * t² = 88.0 m

t² = 88.0 m / 0.250 m/s²

t = 18.8 s

At t = 18.8 s the position of sir George will be

x = 1/2 * 0.300 m/s² * (18.8 s)² = 53.0 m