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Given an acceleration vector, initial velocity $\left(u_{0}, v_{0}, w_{0}\right),\( and initial position \)\left\langle x_{0}, y_{0}, z_{0}\right\rangle,\( find the velocity and position vectors, for \)t \geq 0$. $$\begin{array}{l} \mathbf{a}(t)=\langle 0,0,10\rangle,\left\langle u_{0}, v_{0}, w_{0}\right\rangle=\langle 1,5,0\rangle \\ \left\langle x_{0}, y_{0}, z_{0}\right\rangle=\langle 0,5,0\rangle \end{array}$$

Short Answer

Expert verified
Question: Determine the velocity vector and position vector of an object with an initial velocity vector of (1, 5, 0) and an initial position vector of (0, 5, 0) subjected to a constant acceleration vector of (0, 0, 10). Answer: The velocity vector is \(\mathbf{v}(t) = \langle 1, 5, 10t \rangle\), and the position vector is \(\mathbf{r}(t) = \langle t, 5t + 5, 5t^2 \rangle\) for all \(t \geq 0\).

Step by step solution

01

Integrate acceleration vector to find velocity vector

In order to find the velocity vector of the object, we integrate each component of the acceleration vector with respect to time. So we have: \(u(t) = \int 0 \, dt = 0 + C_1 = 1\), as the initial \(u(t)\) is given by \(u_0 = 1\). \(v(t) = \int 0 \, dt = 0 + C_2 = 5\), as the initial \(v(t)\) is given by \(v_0 = 5\). \(w(t) = \int 10 \, dt = 10t + C_3 = 10t\), as the initial \(w(t)\) is given by \(w_0 = 0\). Thus the velocity vector is: \(\mathbf{v}(t) = \langle 1, 5, 10t \rangle\).
02

Integrate velocity vector to find position vector

Next, we integrate each component of the velocity vector with respect to time. So we have: \(x(t) = \int 1 \, dt = t + C_4 = t\), as the initial \(x(t)\) is given by \(x_0 = 0\). \(y(t) = \int 5 \, dt = 5t + C_5 = 5t + 5\), as the initial \(y(t)\) is given by \(y_0 = 5\). \(z(t) = \int 10t \, dt = \frac{10}{2}t^2 + C_6 = 5t^2\), as the initial \(z(t)\) is given by \(z_0 = 0\). Thus the position vector is: \(\mathbf{r}(t) = \langle t, 5t + 5, 5t^2 \rangle\). Final Answer: The velocity vector is \(\mathbf{v}(t) = \langle 1, 5, 10t \rangle\), and the position vector is \(\mathbf{r}(t) = \langle t, 5t + 5, 5t^2 \rangle\) for all \(t \geq 0\).

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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. \(c(\mathbf{u} \cdot \mathbf{v})=(c \mathbf{u}) \cdot \mathbf{v}=\mathbf{u} \cdot(c \mathbf{v})\)

Find the domains of the following vector-valued functions. $$\mathbf{r}(t)=\sqrt{4-t^{2}} \mathbf{i}+\sqrt{t} \mathbf{j}-\frac{2}{\sqrt{1+t}} \mathbf{k}$$

Determine whether the following statements are true and give an explanation or counterexample. a. The line \(\mathbf{r}(t)=\langle 3,-1,4\rangle+t\langle 6,-2,8\rangle\) passes through the origin. b. Any two nonparallel lines in \(\mathbb{R}^{3}\) intersect. c. The curve \(\mathbf{r}(t)=\left\langle e^{-t}, \sin t,-\cos t\right\rangle\) approaches a circle as \(t \rightarrow \infty\). d. If \(\mathbf{r}(t)=e^{-t^{2}}\langle 1,1,1\rangle\) then \(\lim _{t \rightarrow \infty} \mathbf{r}(t)=\lim _{t \rightarrow-\infty} \mathbf{r}(t)\).

Consider the curve \(\mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k}\) where \(a, b, c, d, e,\) and fare real numbers. It can be shown that this curve lies in a plane. Graph the following curve and describe it. $$\begin{aligned} \mathbf{r}(t)=&(2 \cos t+2 \sin t) \mathbf{i}+(-\cos t+2 \sin t) \mathbf{j} \\\ &+(\cos t-2 \sin t) \mathbf{k} \end{aligned}$$

Find the domains of the following vector-valued functions. $$\mathbf{r}(t)=\sqrt{t+2} \mathbf{i}+\sqrt{2-t} \mathbf{j}$$

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