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Chapter 12: Problem 3

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

Short Answer

Expert verified
Answer: The velocity vector for \(t \geq 0\) is \(\mathbf{v}(t) = \langle 0, 10t + 5 \rangle\), and the position vector is \(\mathbf{r}(t) = \langle 1, 5t^2 + 5t - 1 \rangle\).

Step by step solution

01

Integrate the acceleration vector with respect to time to find the velocity vector

We are given the acceleration vector \(\mathbf{a}(t) = \langle 0, 10 \rangle\). To find the velocity vector, we need to integrate each component of the acceleration vector with respect to time: $$\mathbf{v}(t) = \int \mathbf{a}(t) dt$$ For the x component, we get: $$\int 0 dt = 0t + C_x$$ For the y component, we get: $$\int 10 dt = 10t + C_y$$ So the velocity vector is given by: $$\mathbf{v}(t) = \langle 0t + C_x, 10t + C_y \rangle$$
02

Apply initial velocity conditions to find the constants C_x and C_y

We are given the initial velocity vector as \(\langle u_0, v_0 \rangle = \langle 0, 5 \rangle\). From this, at t=0, we have: $$\mathbf{v}(0) = \langle 0(0) + C_x, 10(0) + C_y \rangle = \langle C_x, C_y \rangle$$ So, we get the constants as \(C_x = 0\) and \(C_y = 5\). Now the velocity vector becomes: $$\mathbf{v}(t) = \langle 0t, 10t + 5 \rangle = \langle 0, 10t + 5 \rangle$$
03

Integrate the velocity vector with respect to time to find the position vector

Now, we need to integrate the velocity vector \(\mathbf{v}(t) = \langle 0, 10t + 5 \rangle\) with respect to time to find the position vector: $$\mathbf{r}(t) = \int \mathbf{v}(t) dt$$ For the x component, we get: $$\int 0 dt = 0t + D_x$$ For the y component, we get: $$\int (10t + 5) dt = 5t^2 + 5t + D_y$$ So the position vector is given by: $$\mathbf{r}(t) = \langle 0t + D_x, 5t^2 + 5t + D_y \rangle$$
04

Apply initial position conditions to find the constants D_x and D_y

We are given the initial position vector as \(\langle x_0, y_0 \rangle = \langle 1, -1 \rangle\). At t=0, we have: $$\mathbf{r}(0) = \langle 0(0) + D_x, 5(0)^2 + 5(0) + D_y \rangle = \langle D_x, D_y \rangle$$ So, we get the constants as \(D_x = 1\) and \(D_y = -1\). Now the position vector becomes: $$\mathbf{r}(t) = \langle 0t + 1, 5t^2 + 5t - 1 \rangle = \langle 1, 5t^2 + 5t - 1 \rangle$$ Thus, for \(t \geq 0\), the velocity vector is \(\mathbf{v}(t) = \langle 0, 10t + 5 \rangle\) and the position vector is \(\mathbf{r}(t) = \langle 1, 5t^2 + 5t - 1 \rangle\).

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Most popular questions from this chapter

An object moves along an ellipse given by the function \(\mathbf{r}(t)=\langle a \cos t, b \sin t\rangle,\) for \(0 \leq t \leq 2 \pi,\) where \(a > 0\) and \(b > 0\) a. Find the velocity and speed of the object in terms of \(a\) and \(b\) for \(0 \leq t \leq 2 \pi\) b. With \(a=1\) and \(b=6,\) graph the speed function, for \(0 \leq t \leq 2 \pi .\) Mark the points on the trajectory at which the speed is a minimum and a maximum. c. Is it true that the object speeds up along the flattest (straightest) parts of the trajectory and slows down where the curves are sharpest? d. For general \(a\) and \(b\), find the ratio of the maximum speed to the minimum speed on the ellipse (in terms of \(a\) and \(b\) ).

Evaluate the following limits. $$\lim _{t \rightarrow 0}\left(\frac{\sin t}{t} \mathbf{i}-\frac{e^{t}-t-1}{t} \mathbf{j}+\frac{\cos t+t^{2} / 2-1}{t^{2}} \mathbf{k}\right)$$

For the following vectors u and \(\mathbf{v}\) express u as the sum \(\mathbf{u}=\mathbf{p}+\mathbf{n},\) where \(\mathbf{p}\) is parallel to \(\mathbf{v}\) and \(\mathbf{n}\) is orthogonal to \(\mathbf{v}\). \(\mathbf{u}=\langle 4,3\rangle, \mathbf{v}=\langle 1,1\rangle\)

In contrast to the proof in Exercise \(81,\) we now use coordinates and position vectors to prove the same result. Without loss of generality, let \(P\left(x_{1}, y_{1}, 0\right)\) and \(Q\left(x_{2}, y_{2}, 0\right)\) be two points in the \(x y\) -plane and let \(R\left(x_{3}, y_{3}, z_{3}\right)\) be a third point, such that \(P, Q,\) and \(R\) do not lie on a line. Consider \(\triangle P Q R\). a. Let \(M_{1}\) be the midpoint of the side \(P Q\). Find the coordinates of \(M_{1}\) and the components of the vector \(\overrightarrow{R M}_{1}\) b. Find the vector \(\overrightarrow{O Z}_{1}\) from the origin to the point \(Z_{1}\) two-thirds of the way along \(\overrightarrow{R M}_{1}\). c. Repeat the calculation of part (b) with the midpoint \(M_{2}\) of \(R Q\) and the vector \(\overrightarrow{P M}_{2}\) to obtain the vector \(\overrightarrow{O Z}_{2}\) d. Repeat the calculation of part (b) with the midpoint \(M_{3}\) of \(P R\) and the vector \(\overline{Q M}_{3}\) to obtain the vector \(\overrightarrow{O Z}_{3}\) e. Conclude that the medians of \(\triangle P Q R\) intersect at a point. Give the coordinates of the point. f. With \(P(2,4,0), Q(4,1,0),\) and \(R(6,3,4),\) find the point at which the medians of \(\triangle P Q R\) intersect.

Torsion formula Show that the formula defining the torsion, \(\tau=-\frac{d \mathbf{B}}{d s} \cdot \mathbf{N},\) is equivalent to \(\tau=-\frac{1}{|\mathbf{v}|} \frac{d \mathbf{B}}{d t} \cdot \mathbf{N} .\) The second formula is generally easier to use.
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