Chapter 13.4, Problem 1PT

To determine

To choose: The appropriate option for the acceleration of the particle whose position at the time t is $3t^{3}i+\ln tj-\sin 2tk$.

Answer to Problem 1PT

The acceleration of the particle is $\underset{\_}{18ti-\frac{1}{t^2}j+4\sin 2tk}$ which is option $\underset{\_}{b}$.

Explanation of Solution

Given that, the position of the given particle at the time t is $r\left(t\right)=3t^{3}i+\ln tj-\sin 2tk$.

Obtain the acceleration for the particle as follows.

Differentiate the above function with respect to t.

$\begin{array}{c}v\left(t\right)=r^{\prime }\left(t\right)\\ =\frac{d}{dt}\left(3t^{3}i+\ln tj-\sin 2tk\right)\\ =\frac{d}{dt}\left(3t^3\right)i+\frac{d}{dt}\left(\ln t\right)j-\frac{d}{dt}\left(\sin 2t\right)k\\ =9t^{2}i+\frac{1}{t}j-2\cos 2tk\end{array}$

Again, differentiate the function $v\left(t\right)$ with respect to t.

$\begin{array}{c}{v}^{\prime }\left(t\right)=r^{″}\left(t\right)\\ =\frac{d}{dt}\left(9t^{2}i+\frac{1}{t}j-2\cos 2tk\right)\\ =\frac{d}{dt}\left(9t^2\right)i+\frac{d}{dt}\left(\frac{1}{t}\right)j-\frac{d}{dt}\left(2\cos 2t\right)k\\ =18ti-\frac{1}{t^2}j+4\sin 2tk\end{array}$

So, the acceleration for the particle at the time t is $18ti-\frac{1}{t^2}j+4\sin 2tk$.

Hence, the acceleration of the particle is $\underset{\_}{18ti-\frac{1}{t^2}j+4\sin 2tk}$ which is option $\underset{\_}{b}$.