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Chapter 12: Problem 2

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 \(t \geq 0\). $$\mathbf{a}(t)=\langle 1,2\rangle,\left\langle u_{0}, v_{0}\right\rangle=(1,1\rangle,\left\langle x_{0}, y_{0}\right\rangle=\langle 2,3\rangle$$

Short Answer

Expert verified
The velocity vector is \(\mathbf{v}(t) = \langle t + 1, 2t + 1 \rangle\) and the position vector is \(\mathbf{r}(t) = \langle \frac{1}{2}t^2 + t + 2, t^2 + t + 3 \rangle\).

Step by step solution

01

Integrate the acceleration vector

To find the velocity vector \(\mathbf{v}(t)\), integrate the acceleration vector \(\mathbf{a}(t)\) with respect to time: $$\mathbf{v}(t) = \int \mathbf{a}(t) dt = \int \langle 1, 2 \rangle dt = \langle \int 1 dt, \int 2 dt \rangle = \langle t + C_1 , 2t + C_2 \rangle$$, where \(C_1\) and \(C_2\) are constants of integration.
02

Apply the initial velocity condition

We are given the initial velocity vector \(\left\langle u_0, v_0 \right\rangle = \langle 1, 1 \rangle\). This gives us two equations to solve for the constants of integration: $$u_0 = t + C_1 \Rightarrow 1 = 0 + C_1 \Rightarrow C_1 = 1$$ $$v_0 = 2t + C_2 \Rightarrow 1 = 0 + C_2 \Rightarrow C_2 = 1$$ Now we can substitute these constants into the velocity vector equation: $$\mathbf{v}(t) = \langle t + 1, 2t + 1 \rangle$$
03

Integrate the velocity vector

To find the position vector \(\mathbf{r}(t)\), integrate the velocity vector \(\mathbf{v}(t)\) with respect to time: $$\mathbf{r}(t) = \int \mathbf{v}(t) dt = \int \langle t + 1, 2t + 1 \rangle dt = \langle \int (t + 1) dt, \int (2t + 1) dt \rangle = \langle \frac{1}{2}t^2 + t + C_3 , t^2 + t + C_4 \rangle$$, where \(C_3\) and \(C_4\) are constants of integration.
04

Apply the initial position condition

We are given the initial position vector \(\left\langle x_0, y_0 \right\rangle = \langle 2, 3 \rangle\). This gives us two equations to solve for the constants of integration: $$x_0 = \frac{1}{2}(0)^2 + 0 + C_3 \Rightarrow 2 = C_3$$ $$y_0 = (0)^2 + 0 + C_4 \Rightarrow 3 = C_4$$ Now we can substitute these constants into the position vector equation: $$\mathbf{r}(t) = \langle \frac{1}{2}t^2 + t + 2, t^2 + t + 3 \rangle$$
05

Final: Velocity and Position Vectors

Combining the information from the previous steps, we find that the velocity vector is: $$\mathbf{v}(t) = \langle t + 1, 2t + 1 \rangle$$ and the position vector is: $$\mathbf{r}(t) = \langle \frac{1}{2}t^2 + t + 2, t^2 + t + 3 \rangle$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acceleration
Understanding acceleration is crucial when discussing velocity and position vectors. Acceleration tells us how quickly the velocity of an object changes over time. It's given as a vector, which in this exercise is \( \mathbf{a}(t) = \langle 1, 2 \rangle \). Each component of this vector represents the change in velocity per unit time in a specific direction鈥攐ften described as horizontal and vertical components on a plane.

Acceleration connects with other concepts like velocity and position through calculus. By finding the velocity vector from acceleration, we perform integration, which is essentially the reverse process of differentiation. It helps in determining how fast and in what direction something is moving at any given point. This rate of change is accumulated over time, beginning at an initial velocity, to find the new velocity and position.
Integration
Integration is the mathematical process used to find the whole from the sum of its parts. Here, we use integration to determine the velocity and position vectors from the given acceleration vector. The acceleration vector \( \mathbf{a}(t) = \langle 1, 2 \rangle \) is integrated with respect to time to find the velocity vector:
  • The integration of \( 1 \) with respect to \( t \) is \( t + C_1 \), where \( C_1 \) is a constant.
  • The integration of \( 2 \) with respect to \( t \) is \( 2t + C_2 \), where \( C_2 \) is another constant.
The calculated velocity vector becomes \( \langle t + C_1 , 2t + C_2 \rangle \).

When integrating the velocity vector to find the position vector, the process is similar:
  • \( \int (t + 1) dt \) gives the result \( \frac{1}{2}t^2 + t + C_3 \).
  • \( \int (2t + 1) dt \) results in \( t^2 + t + C_4 \).
Therefore, the position vector is expressed as \( \langle \frac{1}{2}t^2 + t + C_3 , t^2 + t + C_4 \rangle \).
Initial Conditions
Initial conditions specify the starting point of a system at \( t = 0 \). They are crucial for determining the constants of integration that arise during the integration process. In our problem, we're given initial velocity \( \langle u_0, v_0 \rangle = \langle 1, 1 \rangle \) and position \( \langle x_0, y_0 \rangle = \langle 2, 3 \rangle \).

These conditions allow us to calculate and substitute the values for \( C_1 \), \( C_2 \), \( C_3 \), and \( C_4 \). For example:
  • The condition \( u_0 = 1 \) implies \( C_1 = 1 \).
  • The condition \( v_0 = 1 \) implies \( C_2 = 1 \).
Upon determining the constants, the velocity vector becomes \( \langle t + 1, 2t + 1 \rangle \), and the position vector, using \( x_0 = 2 \) and \( y_0 = 3 \), is \( \langle \frac{1}{2}t^2 + t + 2, t^2 + t + 3 \rangle \). This ensures our solutions match the problem's initial conditions.
Calculus
Calculus is the mathematical study of continuous change and is essential for solving problems involving motion, such as finding velocity and position vectors. It's composed of two main branches: differentiation and integration.

In problems like the one in our exercise, calculus shows us how to model the relationship between position, velocity, and acceleration. Differentiation is used to determine how a quantity changes over time. Conversely, integration accumulates these changes over time to find overall quantities like distance or position.
Through calculus, the concepts of initial velocity and position integrate with the rates given by acceleration, allowing us to construct equations for velocity and position vectors. The power of calculus lies in this ability to transform small, incremental changes into comprehensive solutions for dynamic systems.

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Most popular questions from this chapter

Imagine three unit spheres (radius equal to 1 ) with centers at \(O(0,0,0), P(\sqrt{3},-1,0)\) and \(Q(\sqrt{3}, 1,0) .\) Now place another unit sphere symmetrically on top of these spheres with its center at \(R\) (see figure). a. Find the coordinates of \(R\). (Hint: The distance between the centers of any two spheres is 2.) b. Let \(\mathbf{r}_{i j}\) be the vector from the center of sphere \(i\) to the center of sphere \(j .\) Find \(\mathbf{r}_{O P}, \mathbf{r}_{O Q}, \mathbf{r}_{P Q}, \mathbf{r}_{O R},\) and \(\mathbf{r}_{P R}\).

The points \(P, Q, R,\) and \(S,\) joined by the vectors \(\mathbf{u}, \mathbf{v}, \mathbf{w},\) and \(\mathbf{x},\) are the vertices of a quadrilateral in \(\mathrm{R}^{3}\). The four points needn't lie in \(a\) plane (see figure). Use the following steps to prove that the line segments joining the midpoints of the sides of the quadrilateral form a parallelogram. The proof does not use a coordinate system. a. Use vector addition to show that \(\mathbf{u}+\mathbf{v}=\mathbf{w}+\mathbf{x}\) b. Let \(m\) be the vector that joins the midpoints of \(P Q\) and \(Q R\) Show that \(\mathbf{m}=(\mathbf{u}+\mathbf{v}) / 2\) c. Let n be the vector that joins the midpoints of \(P S\) and \(S R\). Show that \(\mathbf{n}=(\mathbf{x}+\mathbf{w}) / 2\) d. Combine parts (a), (b), and (c) to conclude that \(\mathbf{m}=\mathbf{n}\) e. Explain why part (d) implies that the line segments joining the midpoints of the sides of the quadrilateral form a parallelogram.

An object on an inclined plane does not slide provided the component of the object's weight parallel to the plane \(\left|\mathbf{W}_{\text {par }}\right|\) is less than or equal to the magnitude of the opposing frictional force \(\left|\mathbf{F}_{\mathrm{f}}\right|\). The magnitude of the frictional force, in turn, is proportional to the component of the object's weight perpendicular to the plane \(\left|\mathbf{W}_{\text {perp }}\right|\) (see figure). The constant of proportionality is the coefficient of static friction, \(\mu\) a. Suppose a 100 -lb block rests on a plane that is tilted at an angle of \(\theta=20^{\circ}\) to the horizontal. Find \(\left|\mathbf{W}_{\text {parl }}\right|\) and \(\left|\mathbf{W}_{\text {perp }}\right|\) b. The condition for the block not sliding is \(\left|\mathbf{W}_{\mathrm{par}}\right| \leq \mu\left|\mathbf{W}_{\text {perp }}\right| .\) If \(\mu=0.65,\) does the block slide? c. What is the critical angle above which the block slides with \(\mu=0.65 ?\)

Carry out the following steps to determine the (smallest) distance between the point \(P\) and the line \(\ell\) through the origin. a. Find any vector \(\mathbf{v}\) in the direction of \(\ell\) b. Find the position vector u corresponding to \(P\). c. Find \(\operatorname{proj}_{\mathbf{v}} \mathbf{u}\). d. Show that \(\mathbf{w}=\mathbf{u}-\) projy \(\mathbf{u}\) is a vector orthogonal to \(\mathbf{v}\) whose length is the distance between \(P\) and the line \(\ell\) e. Find \(\mathbf{w}\) and \(|\mathbf{w}| .\) Explain why \(|\mathbf{w}|\) is the distance between \(P\) and \(\ell\). \(P(-12,4) ; \ell: y=2 x\)

a. Show that \((\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+2 \mathbf{u} \cdot \mathbf{v}+|\mathbf{v}|^{2}\). b. Show that \((\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+|\mathbf{v}|^{2}\) if \(\mathbf{u}\) is perpendicular to \(\mathbf{v}\). c. Show that \((\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}-\mathbf{v})=|\mathbf{u}|^{2}-|\mathbf{v}|^{2}\).
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