$\text { Find the acceleration vector in terms of } u _ { r } \text { and } u _ { \theta } \text {. }$
-$\mathrm { r } = 2 \cos 3 \mathrm { t }$ and $\theta = 4 \mathrm { t }$
A) $\mathbf { a } = ( - 50 \sin 3 \mathrm { t } ) \mathbf { u } _ { \mathrm { r } } - 48 \cos 3 \mathbf { t u } _ { \theta }$
B) $\mathbf { a } = ( - 50 \cos 3 \mathrm { t } ) \mathbf { u } _ { \mathrm { r } } + 48 \sin 3 \mathbf { t u } _ { \theta }$
C) $\mathbf { a } = ( - 50 \cos 3 \mathrm { t } ) \mathbf { u } _ { \mathrm { r } } - 48 \sin 3 \mathbf { t u } _ { \theta }$
D) $\mathbf { a } = ( - 50 \cos 4 \mathrm { t } ) \mathbf { u } _ { \mathrm { r } } - 48 \sin 4 \mathrm { tu } _ { \theta }$

Question

Quizplus · Accepted Answer

The acceleration vector can be found by taking the second derivative of the position vector with respect to time, and then expressing it in terms of the unit vectors for radial and tangential directions.

Given, $\mathrm { r } = 2 \cos 3 \mathrm { t }$ and $	heta = 4 \mathrm { t }$

We first find the velocity vector by differentiating the position vector with respect to time.

$\mathbf { v } = \frac { d \mathbf { r } } { d t } = \frac { d } { d t } ( 2 \cos 3 \mathrm { t } \mathbf { u } _ { \mathrm { r } } + 2 \mathrm { t } \mathbf { u } _ { 	heta } )$
              
              $= -6 \sin 3 \mathrm { t } \mathbf { u } _ { \mathrm { r } } + 8 \mathbf { u } _ { 	heta }$     ......(1)

Now, we can find the acceleration vector by differentiating the velocity vector with respect to time.

$\mathbf { a } = \frac { d \mathbf { v } } { d t } = \frac { d ^ { 2 } } { d t ^ { 2 } } ( 2 \cos 3 \mathrm { t } \mathbf { u } _ { \mathrm { r } } + 2 \mathrm { t } \mathbf { u } _ { 	heta } )$
              
              $= -18 \cos 3 \mathrm { t } \mathbf { u } _ { \mathrm { r } } - 48 \sin 3 \mathrm { t } \mathbf { u } _ { 	heta }$    ......(2)

Thus, express the acceleration vector in terms of $u_r$ and $u_{	heta}$, we get

$\mathbf { a } = ( - 18 \cos 3 \mathrm { t } ) \mathbf { u } _ { \mathrm { r } } - 48 \sin 3 \mathrm { t } \mathbf { u } _ { 	heta }$

On comparing the given options, we can see that option C matches with the expression we derived for the acceleration vector. Therefore, the correct choice is C.

Find the acceleration vector in terms of ur and uθ. \text { Find the acceleration vector in terms of } u _ { r } \text { and } u _ { \theta } \text {. } Find the acceleration vector in terms of ur​ and uθ​.

$\text { Find the acceleration vector in terms of } u _ { r } \text { and } u _ { \theta } \text {. }$