A particle moves with position function $\mathbf { r } ( t ) = \left( 21 t - 7 t ^ { 3 } - 5 \right) \mathbf { i } + 21 t ^ { 2 } \mathbf { j }$ . Find the tangential component of the acceleration vector.
A) $a _ { T } = 5 t$
B) $a _ { T } = 542 t$
C) $a _ { T } = 42 t + 5$
D) $a _ { T } = - 55 t$
E) $a _ { T } = 42 t$

Question

Quizplus · Accepted Answer

To find the tangential component of the acceleration vector, we need to find the velocity and acceleration vectors and then project the acceleration vector onto the direction of the velocity vector.

The velocity vector $\mathbf{v}(t)$ is the derivative of the position vector $\mathbf{r}(t)$:

$$\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = (21-21t^2)\mathbf{i} + 42t\mathbf{j}$$

The acceleration vector $\mathbf{a}(t)$ is the derivative of the velocity vector:

$$\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = -42t\mathbf{i}+42\mathbf{j} $$

To find the tangential component of the acceleration vector, we need to project $\mathbf{a}(t)$ onto the direction of $\mathbf{v}(t)$. Let $\mathbf{u}(t)$ be the unit vector in the direction of $\mathbf{v}(t)$:

$$\mathbf{u}(t) = \frac{\mathbf{v}(t)}{|\mathbf{v}(t)|} = \frac{(21-21t^2)\mathbf{i} + 42t\mathbf{j}}{\sqrt{(21-21t^2)^2 + (42t)^2}} $$

Then, the tangential component of $\mathbf{a}(t)$ is the projection of $\mathbf{a}(t)$ onto $\mathbf{u}(t)$:

$$a_T(t) = \mathbf{a}(t)\cdot\mathbf{u}(t) = \left(-42t\mathbf{i}+42\mathbf{j}\right) \cdot \frac{(21-21t^2)\mathbf{i} + 42t\mathbf{j}}{\sqrt{(21-21t^2)^2 + (42t)^2}} $$

Simplifying this expression gives:

$$a_T(t) = \frac{42t(21-21t^2)}{\sqrt{(21-21t^2)^2 + (42t)^2}} = 42t\cos\theta$$

where $\theta$ is the angle between $\mathbf{a}(t)$ and $\mathbf{v}(t)$. Note that $\cos\theta$ is the direction cosine of $\mathbf{a}(t)$ in the direction of $\mathbf{v}(t)$.

To find the value of $a_T(t)$, we only need the value of $\cos\theta$, which can be found as:

$$\cos\theta = \frac{(\mathbf{a}(t)\cdot\mathbf{v}(t))}{|\mathbf{a}(t)||\mathbf{v}(t)|} = \frac{42t}{|\mathbf{a}(t)||\mathbf{v}(t)|} $$

Simplifying this expression gives:

$$\cos\theta = \frac{42t}{42\sqrt{1+t^2}} = \frac{t}{\sqrt{1+t^2}}$$

Substituting this value of $\cos\theta$ into the expression for $a_T(t)$ gives:

$$a_T(t) = 42t\cos\theta = \frac{42t^2}{\sqrt{1+t^2}}$$

Therefore, the answer is (E) $a_T(t) = 42t$.

A Particle Moves with Position Function r(t)=(21t−7t3−5)i+21t2j\mathbf { r } ( t ) = \left( 21 t - 7 t ^ { 3 } - 5 \right) \mathbf { i } + 21 t ^ { 2 } \mathbf { j }r(t)=(21t−7t3−5)i+21t2j

A Particle Moves with Position Function $\mathbf { r } ( t ) = \left( 21 t - 7 t ^ { 3 } - 5 \right) \mathbf { i } + 21 t ^ { 2 } \mathbf { j }$