In a historical movie, two knights on horseback start from rest 93 m apart and ride directly toward each other to do battle. Sir George's accleration has a magnitude of 0.22 m/s2, while Sir Alfred's has a magnitude of 0.34 m/s2. How far from Sir George's starting point do the knights collide?
d1 + d2 = 93 m.
0.5a1*t^2 + 0.5a2*t^2 = 93,
0.11*t^2 + 0.17*t^2 = 93,
0.28t^2 = 93,
t^2 = 332.1,
t = 18.2 s. = Time elapsed before collision,
d1 = 0.11*(18.2)^2 = 36.4 m.= Distance
from Sir George's starting point.
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To find the distance from Sir George's starting point where the knights collide, we need to determine the time it takes for them to collide. Then, we can use this time to calculate Sir George's displacement.
Let's start by finding the time it takes for the knights to collide. We can use the second equation of motion, which relates displacement, initial velocity, acceleration, and time:
s = ut + (1/2)at^2
For Sir George:
Initial velocity (u) = 0 m/s (starting from rest)
Acceleration (a) = 0.22 m/s^2
Displacement (s) = ?
Since we want to find the distance from Sir George's starting point, we can let his displacement be represented by "s" in the equation. Therefore, the equation becomes:
s = (1/2)at^2
Next, let's find the time it takes for the knights to collide. Since they are approaching each other, their relative velocity would be the sum of their individual velocities.
Relative velocity = Sir George's velocity + Sir Alfred's velocity
The initial velocities for both knights are 0, so we can ignore them for finding the relative velocity.
Relative acceleration = Sir George's acceleration + Sir Alfred's acceleration
= 0.22 m/s^2 + 0.34 m/s^2
= 0.56 m/s^2
Now, let's use the equation of motion again to find the time (t) it takes for the knights to collide:
93 m = (1/2)(0.56 m/s^2)t^2
Rearranging the equation:
t^2 = (2 * 93 m) / (0.56 m/s^2)
t^2 = 332.14 s^2
Taking the square root of both sides:
t = sqrt(332.14 s^2)
t = 18.22 s (rounded to two decimal places)
Now that we have the time it takes for the knights to collide, we can find Sir George's displacement by substituting this time (t) into the first equation of motion:
s = ut + (1/2)at^2
For Sir George:
Initial velocity (u) = 0 m/s (starting from rest)
Acceleration (a) = 0.22 m/s^2
Time (t) = 18.22 s
Substituting the values into the equation:
s = 0 m/s * 18.22 s + (1/2)(0.22 m/s^2)(18.22 s)^2
s = 0 + 0.22 m/s^2 * 166.2 s^2
s = 36.56 m (rounded to two decimal places)
Therefore, the knights collide with Sir George at a distance of approximately 36.56 meters from his starting point.