Find the scalar tangential and normal components of acceleration of a particle with position vector $\mathbf { r } ( t ) = e ^ { t } \{ \cos 6 t , \sin 6 t , 0 \}$
A) $a _ { \mathbf { T } } = \sqrt { 37 } e ^ { t } , a _ { \mathrm { N } } = 6 \sqrt { 37 } e ^ { t }$
B) $a _ { \mathbf { T } } = 6 e ^ { t } , a _ { \mathbf { N } } = 37 e ^ { t }$
C) $a _ { \mathbf { T } } = 6 \sqrt { 37 } e ^ { t } , a _ { \mathrm { N } } = \sqrt { 37 } e ^ { t }$
D) $a _ { \mathbf { T } } = 37 e ^ { t } , a _ { \mathrm { N } } = 6 e ^ { t }$

Question

Quizplus · Accepted Answer

The scalar tangential component of acceleration is given by $$a_T = \sqrt{a_x^2 + a_y^2}$$where $a_x$ and $a_y$ are the components of acceleration in the x and y directions, respectively.The scalar normal component of acceleration is given by $$a_N = \frac{|\mathbf{a} \times \mathbf{v}|}{|\mathbf{v}|}$$where $\mathbf{a}$ is the acceleration vector and $\mathbf{v}$ is the velocity vector.For the given position vector, we have$$\mathbf{r}(t) = e^t \{\cos 6t, \sin 6t, 0\}$$The velocity vector is given by$$\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = e^t \{-\sin 6t, 6\cos 6t, 0\}$$The acceleration vector is given by$$\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = e^t \{-6\cos 6t, -36\sin 6t, 0\}$$Therefore, the scalar tangential component of acceleration is$$a_T = \sqrt{(-6\cos 6t)^2 + (-36\sin 6t)^2} = 6\sqrt{37}e^t$$The scalar normal component of acceleration is$$a_N = \frac{|\mathbf{a} \times \mathbf{v}|}{|\mathbf{v}|} = \frac{e^t \sqrt{(6\cos 6t)(6\cos 6t) + (36\sin 6t)(36\sin 6t)}}{e^t \sqrt{(-\sin 6t)^2 + (6\cos 6t)^2}} = \sqrt{37}e^t$$

Find the Scalar Tangential and Normal Components of Acceleration of a Particle