94=1/2 .22*t^2+1/2 .27 t^2
solve for t, the time of collision.
Then use that t to solve for distance traversed.
solve for t, the time of collision.
Then use that t to solve for distance traversed.
d=???
Let's start by calculating the time it takes for each knight to collide. We can use the following kinematic equation:
distance = initial velocity × time + (0.5 × acceleration × time^2)
For Sir George:
distance = initial velocity × time + (0.5 × acceleration × time^2)
94m = 0 × t + (0.5 × 0.22 m/s^2 × t^2)
94m = 0.11 m/s^2 × t^2
For Sir Alfred:
distance = initial velocity × time + (0.5 × acceleration × time^2)
94m = 0 × t + (0.5 × 0.27 m/s^2 × t^2)
94m = 0.135 m/s^2 × t^2
Now, we have two equations:
Equation 1: 94 = 0.11t^2
Equation 2: 94 = 0.135t^2
Solving these equations, we find that t ≈ 12.520 seconds for Equation 1 and t ≈ 11.205 seconds for Equation 2.
Now that we know the time it takes for each knight to collide, we can find the distance they've traveled from Sir George's starting point.
For Sir George:
distance = initial velocity × time + (0.5 × acceleration × time^2)
distance = 0 × 12.520s + (0.5 × 0.22 m/s^2 × 12.520s^2)
distance ≈ 0.1376 m
For Sir Alfred:
distance = initial velocity × time + (0.5 × acceleration × time^2)
distance = 0 × 11.205s + (0.5 × 0.27 m/s^2 × 11.205s^2)
distance ≈ 0.1686 m
Therefore, Sir George and Sir Alfred would collide approximately 0.1376m away from Sir George's starting point.