Question

the maximum acceleration attained on the interval ( 0 leq t leq 3 ) by the particle whose velocity is given by ( v(t)=t^{3}-3 t^{2}+12 t+4 ) is

Explanation:

Step1: Find the acceleration function

Acceleration \(a(t)\) is the derivative of velocity \(v(t)\). Given \(v(t)=t^{3}-3t^{2}+12t + 4\), using the power rule \(\frac{d}{dt}(x^{n})=nx^{n - 1}\), we have \(a(t)=v^{\prime}(t)=3t^{2}-6t + 12\).

Step2: Find the derivative of the acceleration function

To find the extremum of \(a(t)\), we find its derivative \(a^{\prime}(t)\). Using the power rule again, \(a^{\prime}(t)=(3t^{2}-6t + 12)^{\prime}=6t-6\).

Step3: Find the critical points

Set \(a^{\prime}(t) = 0\), so \(6t-6=0\). Solving for \(t\), we get \(t = 1\).

Step4: Check the endpoints and the critical point

We evaluate \(a(t)\) at the endpoints \(t = 0\) and \(t=3\) and the critical point \(t = 1\).

• When \(t = 0\), \(a(0)=3(0)^{2}-6(0)+12=12\).
• When \(t = 1\), \(a(1)=3(1)^{2}-6(1)+12=3 - 6+12=9\).
• When \(t = 3\), \(a(3)=3(3)^{2}-6(3)+12=27-18 + 12=21\).

Answer:

\(21\) (corresponds to option D)