Let the position function of a particle be $\mathbf { r } ( t ) = t \mathbf { i } + t ^ { 2 } \mathbf { j }$ . Find the normal component of the acceleration vector when t = 1.
A) $\frac { 1 } { \sqrt { 5 } }$

B) $\frac { 2 } { \sqrt { 5 } }$
C) $\frac { 3 } { \sqrt { 5 } }$
D) $\frac { 4 } { \sqrt { 5 } }$
E) $\sqrt { 5 }$
F) $\frac { 6 } { \sqrt { 5 } }$

G) $\frac { 7 } { \sqrt { 5 } }$

H) $\frac { 8 } { \sqrt { 5 } }$

Question

Let the position function of a particle be $\mathbf { r } ( t ) = t \mathbf { i } + t ^ { 2 } \mathbf { j }$ . Find the normal component of the acceleration vector when t = 1.
A) $\frac { 1 } { \sqrt { 5 } }$ 
 
B) $\frac { 2 } { \sqrt { 5 } }$ 
C) $\frac { 3 } { \sqrt { 5 } }$ 
D) $\frac { 4 } { \sqrt { 5 } }$ 
E) $\sqrt { 5 }$
F) $\frac { 6 } { \sqrt { 5 } }$

G) $\frac { 7 } { \sqrt { 5 } }$

H) $\frac { 8 } { \sqrt { 5 } }$

Quizplus · Accepted Answer

The normal component of acceleration is given by $ a_N = \frac{| \mathbf{v} \times \mathbf{a} |}{|\mathbf{v}|} $. First, find the velocity $\mathbf{v}(t) = \mathbf{i} + 2t\mathbf{j}$ and acceleration $\mathbf{a}(t) = 2\mathbf{j}$. At $t = 1$, $\mathbf{v}(1) = \mathbf{i} + 2\mathbf{j}$ and $\mathbf{a}(1) = 2\mathbf{j}$. The cross product $|\mathbf{v} \times \mathbf{a}|$ is 2, and $|\mathbf{v}|$ is $\sqrt{5}$. Thus, $a_N = \frac{2}{\sqrt{5}}$.

Let the Position Function of a Particle Be r(t)=ti+t2j\mathbf { r } ( t ) = t \mathbf { i } + t ^ { 2 } \mathbf { j }r(t)=ti+t2j

Let the Position Function of a Particle Be $\mathbf { r } ( t ) = t \mathbf { i } + t ^ { 2 } \mathbf { j }$