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Chapter 12: Problem 35

Consider the following trajectories of moving objects. Find the tangential and normal components of the acceleration. $$\mathbf{r}(t)=\langle t, 1+4 t, 2-6 t\rangle$$

Short Answer

Expert verified
Answer: The tangent acceleration component \(a_t\) is 0 and the normal acceleration component \(a_n\) is 0 for the given trajectory.

Step by step solution

01

Find the velocity vector

To find the velocity vector \(\mathbf{v}(t)\), we need to differentiate the position vector \(\mathbf{r}(t)\) with respect to time: $$\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt}$$ So we differentiate each component of \(\mathbf{r}(t)\): $$\mathbf{v}(t) = \left\langle \frac{d}{dt}(t), \frac{d}{dt}(1+4t), \frac{d}{dt}(2-6t) \right\rangle = \langle 1, 4, -6 \rangle$$
02

Find the acceleration vector

To find the acceleration vector \(\mathbf{a}(t)\), we need to differentiate the velocity vector \(\mathbf{v}(t)\) with respect to time: $$\mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt}$$ So we differentiate each component of \(\mathbf{v}(t)\): $$\mathbf{a}(t) = \left\langle \frac{d}{dt}(1), \frac{d}{dt}(4), \frac{d}{dt}(-6) \right\rangle = \langle 0, 0, 0 \rangle$$ Now we have the acceleration vector, which is a zero vector.
03

Find the tangent acceleration component

Tangent acceleration component is the projection of the acceleration vector onto the velocity vector: $$a_t = \frac{\mathbf{a} \cdot \mathbf{v}}{\lVert\mathbf{v}\rVert}$$ However, since the acceleration vector is a zero vector, the tangent acceleration component is 0.
04

Find the normal acceleration component

The normal acceleration component can be found using the following formula: $$a_n = \sqrt{\lVert\mathbf{a}\rVert^2 - a_t^2}$$ Since the acceleration vector is a zero vector, its magnitude is also 0, and the normal acceleration component is also 0.
05

Conclusion

For the given trajectory, the tangent acceleration component \(a_t\) is 0 and the normal acceleration component \(a_n\) is 0 due to the constant velocity vector and zero acceleration vector.

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Most popular questions from this chapter

Let \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) \(\mathbf{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle\), and \(\mathbf{w}=\) \(\left\langle w_{1}, w_{2}, w_{3}\right\rangle\). Let \(c\) be a scalar. Prove the following vector properties. \(\mathbf{u} \cdot(\mathbf{v}+\mathbf{w})=\mathbf{u} \cdot \mathbf{v}+\mathbf{u} \cdot \mathbf{w}\)

Evaluate the following limits. $$\lim _{t \rightarrow \pi / 2}\left(\cos 2 t \mathbf{i}-4 \sin t \mathbf{j}+\frac{2 t}{\pi} \mathbf{k}\right)$$

The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|(\text {because}|\cos \theta| \leq 1) .\) This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. What conditions on \(\mathbf{u}\) and \(\mathbf{v}\) lead to equality in the Cauchy-Schwarz Inequality?

Let \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) \(\mathbf{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle\), and \(\mathbf{w}=\) \(\left\langle w_{1}, w_{2}, w_{3}\right\rangle\). Let \(c\) be a scalar. Prove the following vector properties. \(c(\mathbf{u} \cdot \mathbf{v})=(c \mathbf{u}) \cdot \mathbf{v}=\mathbf{u} \cdot(c \mathbf{v})\)

Find the point (if it exists) at which the following planes and lines intersect. $$z=-8 ; \mathbf{r}(t)=\langle 3 t-2, t-6,-2 t+4\rangle$$
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