Answer:
D
Step-by-step explanation:
Sophia is more correct in her solution.
Apply slip and slide
a^2-3a-4
(a-4)(a+1)
(a-2)(2a+1)
Find zeros
a-2=0
a=2
2a+1=0
2a=-1
a=-1/2
Final answer: {-1/2, 2}
<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
2x - (3x + 2x)
= -3x
hope this helps !!
