Chapter 11.2, Problem 55E

To determine

Todetermine:Thecorrect option for acceleration vector of the particle when $x=t^2+1$ and $y=\ln \left(2t+3\right)$ for all $t\ge 0$ from $\left(\text{A}\right)\left(2t,\frac{2}{2t+3}\right)$

  $\left(\text{B}\right)\left(2t,-\frac{4}{{\left(2t+3\right)}^2}\right)$

  $\left(\text{C}\right)\left(2,\frac{4}{{\left(2t+3\right)}^2}\right)$

  $\left(\text{D}\right)\left(2,\frac{2}{{\left(2t+3\right)}^2}\right)$

  $\left(\text{E}\right)\left(2,-\frac{4}{{\left(2t+3\right)}^2}\right)$ .

Answer to Problem 55E

The correct option is (E).

Explanation of Solution

Given information:

  $x=t^2+1$

  $y=\ln \left(2t+3\right)$

  $t\ge 0$

Formula Used:

Acceleration vector of an object with position vector $\langle x\left(t\right),y\left(t\right)\rangle$ at time $t$ is:

  $a\left(t\right)=\langle \frac{d^2}{d{t}^2}x\left(t\right),\frac{d^2}{d{t}^2}y\left(t\right)\rangle$

Calculation:

The given position vector is $\langle x,y\rangle$ .

Now, calculate the acceleration vector. Use the formula $a\left(t\right)=\langle \frac{d^2}{d{t}^2}x\left(t\right),\frac{d^2}{d{t}^2}y\left(t\right)\rangle$ .

  $\begin{array}{c}a\left(t\right)=\frac{d^2}{d{t}^2}\langle x,y\rangle \\ =\frac{d^2}{d{t}^2}\langle t^2+1,\ln \left(2t+3\right)\rangle \\ =\langle \frac{d^2}{d{t}^2}\left(t^2+1\right),\frac{d^2}{d{t}^2}\left(\ln \left(2t+3\right)\right)\rangle \\ =\langle 2,-\frac{4}{{\left(2t+3\right)}^2}\rangle \end{array}$

Thus, the acceleration vector $a\left(t\right)$ is $\langle 2,-\frac{4}{{\left(2t+3\right)}^2}\rangle$ .

Conclusion:

Hence, the correct option is (E).