So hmmm check the picture below
so, we're looking for dr/dt then at 4:00pm or 4 hours later
now, keep in mind that, the distance "x", is not changing, is constant whilst "y" and "r" are moving, that simply means when taking the derivative, that goes to 0
The answer is d
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Answer:
work is attached and shown
<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