How far from Sir George’s starting point do the knights collide?

Cindy asked:


Two knights on horseback start from rest 77m apart and ride directly toward each other to do battle. Sir George’s acceleration has a magnitude of 0.15 m/s^2, while Sir Alfred’s has a magnitude of 0.34 m/s^2.

horseback

2 Responses to “How far from Sir George’s starting point do the knights collide?”

1. tinfoil666 says:

   June 19, 2009 at 11:21 pm

   horseback

   General displacement equation:
   x(t) = 0.5 *a*t^2 + Vo(t) + Xo.

   Initial velocity (Vo) = zero for both.

   Sir George’s displacement equation (assuming his starting point is defined as ‘zero’):
   x1(t) = 0.15*0.5 m/s * t^2 + 0 + 0

   opponent’s equation:
   x2(t) = -0.34 *0.5 m/s * T^2 + 0 + 77m

   When they meet, x1=x2.
   Two equations, two unknowns (x1 and t). Solve away.
2. kuiperbelt2003 says:

   June 16, 2009 at 7:34 pm

   horseback

   the general equation of motion is:

   x(t)=x0+v0t+1/2at^2

   where x(t) is the position at any time t, x0 is the initial position, v0 is the initial speed, a is the acceleration, and t is the elapsed time

   for these two riders, their equations become:

   x(george)=0+0+1/2(0.15)t^2
   x(alfred)=77+0-1/2(0.34)t^2

   they meet when the two expressions are equal:

   1/2(0.15)t^2=77-1/2(0.34)t^2

   or

   0.245t^2=77 -> t=17.73s

   at t=17.73 secs, george has traveled x=1/2(0.15)(17.73)^2
   = 23.5m, so they meet 23.5 meters from george’s starting spot