So I have the formula:
$v(t) = 0.001302t^3 - 0.09029t^2 +23.61t - 3.083$
Determine the minimum and maximum acceleration in the interval: $t ∈ [0,126] s$
So I know that $v'(t) = a(t)$ so $a(t) = 3(0.001302)t^2 - 2(0.09029)t + 23.61$
So now I know that I can find the minimum acceleration at $a'(t) = 0$ and solving that I find that $t < 0$ so that tells me the minimum is at 0s which I can plug in to find minimum, and because of that I know that maximum is at 126s.
Is my reasoning correct?
1 Answer
$a(t)=v'(t)=0.003906 t^2-0.18058 t+23.61$
The acceleration is minimum when $a'(t)=0\land a''(t)>0$
$a'(t)=0.007812 t-0.18058$
$a'(t)=0$ for $t_0=23.1157$s
$a''(t)=0.007812>0$ thus $t_0$ is a local minimum
as the derivative $a(t)$ is positive for $t>t_0$ the acceleration is increasing so the minimum is also a global minimum.
Furthermore we can observe that $a(t)$ is the equation of a parabola with concavity up, therefore the local minimum is also the global minimum
Hope this helps
