Question

In a historical movie, two knights on horseback start from rest 97 m apart and ride directly toward each other to do battle. sir george's acceleration has a magnitude of 0.16 m/s2, while sir alfred's has a magnitude of 0.30 m/s2. relative to sir george's starting point, where do the knights collide?

Answer 1

The situation is equivalent with one knight standing and the other moving with the sum of the two accelerations. Thus we can find the time from the equation
$s=(a_1+a_2)t^2/2$
$t= \sqrt{2s/(a_1+a_2)} = \sqrt{2*97/(0.16+0.30)} = 20.536 s$

Relative to Sir George stating point the distance where they encounter is
$x=a_1t^2/2 =0.16*20.536^2/2 =33.74 (meters)$

Answer 2

Let t = the time before the two knights collide.

The distance traveled by Sir George is
s₁ = (1/2)*(0.16 m/s²)*(t s)² = 0.08t² m

The distance traveled by Sir Alfred is
s₂ = (1/2*(0.30 m/s²)*(t s)² = 0.15t² m

The total distance should equal 97 m, therefore
0.08t² + 0.15t² = 97
0.23t² = 97
t = 20.5363 s

The distance that Sir George travels is
s₁ = 0.08*20.5363² = 33.74 m

Answer: 33.74 m