for time t > 0, the position of a particle moving in the xy - plane is given by the parametric equations x =…

Answer:

B. $\left(2,\frac{9}{32}\right)$

Explanation:

Step1: Find the first - derivative of $x$ with respect to $t$

Differentiate $x = 4t+t^{2}$ using the power rule. $\frac{dx}{dt}=4 + 2t$.

Step2: Find the second - derivative of $x$ with respect to $t$

Differentiate $\frac{dx}{dt}=4 + 2t$ with respect to $t$. $\frac{d^{2}x}{dt^{2}}=2$.

Step3: Find the first - derivative of $y$ with respect to $t$

Use the quotient rule. If $y=\frac{1}{3t + 1}=(3t + 1)^{-1}$, then $\frac{dy}{dt}=-1\times(3t + 1)^{-2}\times3=-\frac{3}{(3t + 1)^{2}}$.

Step4: Find the second - derivative of $y$ with respect to $t$

Use the quotient rule or chain rule. $\frac{d^{2}y}{dt^{2}}=\frac{3\times2\times3}{(3t + 1)^{3}}=\frac{18}{(3t + 1)^{3}}$.

Step5: Evaluate the second - derivatives at $t = 1$

For $x$, $\frac{d^{2}x}{dt^{2}}\big|{t = 1}=2$. For $y$, $\frac{d^{2}y}{dt^{2}}\big|{t = 1}=\frac{18}{(3\times1+ 1)^{3}}=\frac{18}{64}=\frac{9}{32}$. The acceleration vector is $\left(\frac{d^{2}x}{dt^{2}},\frac{d^{2}y}{dt^{2}}\right)=\left(2,\frac{9}{32}\right)$ at $t = 1$.