A) AB is where plane P and plane R intersect.
B) A, D, B are collinear (on the same line).
C) plane ADG (three points on the plane)
D) F, D, G, A are all on plane R
E) D lies on both planes.
H=0
Therefore:
144t-16t²=0
t (144-16t)=0
We have two equations:
1)
t=0 (when the ball is released)
2)
144-16t=0
-16t=-144
t=-144 / -16=9 (this is the time )
Answer: the ball hit the ground in 9 s.
<span>the particle's initial position is at t=0, x = 0 - 0 + 4 = 4m
velocity is rate of change of displacement = dx/dt = d(t^3 - 9t^2 +4)/dt
= 3t^2 - 18t
acceleration is rate of change of velocity = d(3t^2 -18t)/dt
= 6t - 18
</span><span>the particle is stationary when velocity = 0, so 3t^2 - 18t =0
</span>3t*(t - 6) = 0
t = 0 or t = 6s
acceleration = 6t - 18 = 0
t = 3s
at t = 3s, velocity = 3(3^2) -18*3 = -27m/s
displacement = 3^3 - 9*3^2 +4 = -50m
