In an historical movie, two knights on horseback start from rest 90.6 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.340 m/s2, while Sir Alfred's has a magnitude of 0.346 m/s2. Relative to Sir George's starting point, where do the knights collide?
To find the point where the knights collide, we need to determine the time it takes for them to meet. We can do this by using kinematic equations.
First, let's find the time it takes for Sir George to reach the point of collision. We can use the following equation:
s = ut + (1/2)at^2
Where:
- s is the distance traveled
- u is the initial velocity (which is 0 since Sir George starts from rest)
- a is the acceleration
- t is the time
For Sir George:
s1 = ut + (1/2)at^2
Since he is starting from rest, u1 = 0. The distance s1 is the distance between Sir George's starting point and the point of collision. Let's call this distance d.
d = 0t + (1/2)at^2
d = (1/2)at^2
Now, let's find the time it takes for Sir Alfred to reach the point of collision using the same equation:
s2 = ut + (1/2)at^2
The distance s2 is the distance between Sir Alfred's starting point and the point of collision. Since they start 90.6 m apart, s2 = 90.6 - d.
s2 = (90.6 - d)
s2 = ut + (1/2)at^2
90.6 - d = (1/2)at^2
Now we have two equations with two unknowns (d and t). Let's solve these equations simultaneously to find the values of d and t.
Equation 1: d = (1/2)at^2
Equation 2: 90.6 - d = (1/2)at^2
Substituting Equation 1 into Equation 2:
90.6 - (1/2)at^2 = (1/2)at^2
Simplifying:
90.6 = (1/2)at^2 + (1/2)at^2
90.6 = at^2
Now we can solve for t:
t^2 = 90.6 / a
t = √(90.6 / a)
Plug in the values of a for Sir George:
t = √(90.6 / 0.340)
t = √266.47
t ≈ 16.34 seconds
Now that we have the time, let's find the distance d:
d = (1/2)at^2
d = (1/2)(0.340)(16.34)^2
d ≈ 93.67 m
The knights collide at a distance of approximately 93.67 meters relative to Sir George's starting point.