Two knights on horseback start from rest 88.0 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of .300 m/s^2, while Sir Alfred's has a magnitude of .200 m/s^2. Relative to sir Ggeorge's starting point, where do the knights collide?
.150t^2 + .100t^2 = 88.0
the two distances are in the ratio 3:2
so, ...
a boy pushed a shopping cart with a force of 10N, and the cart accelerates 1m/s2 . what is the mass of the cart
To find the collision point, we can first determine the time it takes for the two knights to collide by using kinematic equations. We'll use the equation:
Δx = v_0t + (1/2)at^2
For Sir George:
Initial velocity v_0 = 0 m/s (starting from rest)
Acceleration a = 0.300 m/s^2
Distance Δx = (distance between the two knights) / 2 = 88.0 m / 2 = 44.0 m
Plugging in the values:
44.0 m = (1/2)(0)(t^2) + (1/2)(0.300 m/s^2)(t^2)
Simplifying the equation:
44.0 m = 0.150 m/s^2 * t^2
t^2 = (44.0 m) / (0.150 m/s^2)
t = √(293.333 s^2)
t ≈ 17.123 s
So, it takes approximately 17.123 seconds for Sir George to reach the collision point.
Now, let's find out how far Sir George travels in that time:
Using the equation:
Δx = v_0t + (1/2)at^2
Initial velocity v_0 = 0 m/s (starting from rest)
Acceleration a = 0.300 m/s^2
Time t = 17.123 s
Plugging in the values:
Δx = (1/2)(0)(17.123 s) + (1/2)(0.300 m/s^2)(17.123 s)^2
Simplifying the equation:
Δx = 0 + (0.150 m/s^2)(292.998 s^2)
Δx = 43.950 m
Therefore, relative to Sir George's starting point, the knights collide approximately 43.950 meters away from Sir George's starting point.