In a historical movie, two knights on horseback start from rest 94m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.22 m/s2, while Sir Alfred's has a magnitude of 0.27 m/s2. Relative to Sir George's starting point, where do the knights collide?

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.

t=6.85seconds

d=???

To determine where the knights collide relative to Sir George's starting point, we first need to calculate the time it takes for each knight to collide. Once we have the time, we can use it to find the distance each knight travels before the collision.

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.