(t^3)+(t^2)-16t+12 for 0<=t<=8 describe the particles motion when t=2
can u please tell me about it's maximum and its acceleration whether it speeding up, or slowing down, or whether its constant
I know that it has a max of -8 but i don't know how they got it please help.
To describe the particle's motion when t=2, we need to find the position of the particle at that time. We can do this by substituting t=2 into the given equation:
(t^3) + (t^2) - 16t + 12
= (2^3) + (2^2) - 16(2) + 12
= 8 + 4 - 32 + 12
= -8
Therefore, when t=2, the position of the particle is -8.
To find the maximum value, we can determine the vertex of the cubic function. The vertex of a cubic function in the form of f(x) = ax^3 + bx^2 + cx + d is given by the formula:
x = -b / (3a)
In our case, a=1, b=1, and c=-16, so x = -1 / (3*1) = -1/3.
Now, substitute this value back into the original equation to find the maximum value:
f(-1/3) = ((-1/3)^3) + ((-1/3)^2) - 16(-1/3) + 12
= (-1/27) + (1/9) + (16/3) + 12
≈ -0.037
Hence, the maximum value is approximately -0.037.
To determine whether the particle is speeding up, slowing down, or moving at a constant speed, we can examine the particle's acceleration. The acceleration is given by the derivative of the position function. Let's differentiate the given equation to find the acceleration function:
a(t) = d^2/dt^2 [(t^3) + (t^2) - 16t + 12]
= 6t + 2
Now, substitute t=2 into the acceleration function to find the particle's acceleration at t=2:
a(2) = 6(2) + 2
= 14
Since the acceleration is positive (a(2) = 14), the particle is speeding up when t=2.