angle billiards pool
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Two billiard balls are moving on a pool table according to the respective paths given below, where t represent?
measured in seconds.
x(t)=(t^2-6, t^2/2 – 14) and y(t)=(5t, 40-t^2)
a. when and where do the balls collide?
b. what is the angle formed by the paths of the balls at the collision point ?
The first coordinates cross when
t^2 – 6 = 5t
(t-6)(t+1) = 0
t=+6 or t=-1
The second coordinates cross when
t^2/2 – 14 = 40-t^2
3t^2 -108 = 0
t^2 = 36
t=+6 or t=-6
To hit, both coordinates have to be equal at the same time. So apparently, the balls hit at t=+6. Plug that into either x(t) or y(t) to get the collision point: (30,4)
To get the slopes, take derivatives
x’(t) = (2t,t)
y’(t) = (5, -2t)
x’(6) = (12,6),
so the slope for that ball is 6/12 = 1/2
and the angle is arctan(1/2)
y’(6) = (5,-12)
so the slope is -12/5
and the angle is the arctangent of that.
Take the difference of the angles.
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