Find the scalar tangential and normal components of acceleration of a particle with position vector $\mathbf { r } ( t ) = 3 \sin t \mathbf { i } + 3 \cos t \mathbf { j } + 5 t \mathbf { k }$
A) $\quad a _ { \mathbf { T } } = 0 , a _ { \mathrm { N } } = 3$
B) $a _ { \mathbf { T } } = 5 , a _ { \mathbf { N } } = 3$
C) $a _ { \mathbf { T } } = 0 , a _ { \mathrm { N } } = 3 \sqrt { 26 }$
D) $a _ { \mathbf { T } } = 5 , a _ { \mathrm { N } } = 3 \sqrt { 26 }$

Question

Quizplus · Accepted Answer

The scalar tangential component of acceleration is given by $$a_T = \frac{d}{dt} \left[ \frac{1}{2} \frac{d\mathbf{r}}{dt} \cdot \frac{d\mathbf{r}}{dt} \right] = \frac{1}{2} \frac{d}{dt} \left[ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2 \right]$$$$= \frac{1}{2} \left[ 2\frac{dx}{dt} \frac{d^2x}{dt^2} + 2\frac{dy}{dt} \frac{d^2y}{dt^2} + 2\frac{dz}{dt} \frac{d^2z}{dt^2} \right]$$$$= \frac{dx}{dt} \frac{d^2x}{dt^2} + \frac{dy}{dt} \frac{d^2y}{dt^2} + \frac{dz}{dt} \frac{d^2z}{dt^2}$$$$= 3\cos t (-3\sin t) + 3\sin t (3\cos t) + 5(0) = 0$$The scalar normal component of acceleration is given by $$a_N = \sqrt{ \left( \frac{d^2x}{dt^2} \right)^2 + \left( \frac{d^2y}{dt^2} \right)^2 + \left( \frac{d^2z}{dt^2} \right)^2 - a_T^2 }$$$$= \sqrt{ (-3\sin t)^2 + (3\cos t)^2 - 0^2 } = 3$$

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