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Use the given information to find the position and velocity vectors of the particle. $$ \mathbf{a}(t)=\mathbf{i}+e^{-t} \mathbf{j} ; \mathbf{v}(0)=2 \mathbf{i}+\mathbf{j} ; \mathbf{r}(0)=\mathbf{i}-\mathbf{j} $$

Short Answer

Expert verified
The velocity vector is \( \mathbf{v}(t) = (t + 2)\mathbf{i} + (-e^{-t} + 2)\mathbf{j} \); the position vector is \( \mathbf{r}(t) = \left(\frac{t^2}{2} + 2t + 1\right)\mathbf{i} + (e^{-t} + 2t - 2)\mathbf{j} \).

Step by step solution

01

Find the velocity vector

The acceleration vector is given by \( \mathbf{a}(t) = \mathbf{i} + e^{-t} \mathbf{j} \). To find the velocity vector, we need to integrate the acceleration vector. Let \( \mathbf{v}(t) = \int \mathbf{a}(t) \, dt \). Integrating each component, we get: \( v_x(t) = \int 1 \, dt = t + C_1 \) for the \( \mathbf{i} \) component, and \( v_y(t) = \int e^{-t} \, dt = -e^{-t} + C_2 \) for the \( \mathbf{j} \) component. So, \( \mathbf{v}(t) = (t + C_1)\mathbf{i} + (-e^{-t} + C_2)\mathbf{j} \).
02

Apply initial condition to find constants

We know \( \mathbf{v}(0) = 2\mathbf{i} + \mathbf{j} \). Setting \( t = 0 \) in the velocity expression, we get \( \mathbf{v}(0) = (0 + C_1)\mathbf{i} + (-e^0 + C_2)\mathbf{j} = C_1\mathbf{i} + (C_2 - 1)\mathbf{j} \). Given \( C_1\mathbf{i} + (C_2 - 1)\mathbf{j} = 2\mathbf{i} + \mathbf{j} \), we compare the components to find \( C_1 = 2 \) and \( C_2 - 1 = 1 \). Solving gives \( C_2 = 2 \).
03

Construct the velocity vector

Substitute the constants \( C_1 = 2 \) and \( C_2 = 2 \) back into the expression for \( \mathbf{v}(t) \). This gives \( \mathbf{v}(t) = (t + 2)\mathbf{i} + (-e^{-t} + 2)\mathbf{j} \).
04

Find the position vector

With \( \mathbf{v}(t) \) known, find the position vector \( \mathbf{r}(t) \) by integrating \( \mathbf{v}(t) \). Let \( \mathbf{r}(t) = \int \mathbf{v}(t) \, dt \). For the \( \mathbf{i} \) component: \( r_x(t) = \int (t + 2) \, dt = \frac{t^2}{2} + 2t + C_3 \). For the \( \mathbf{j} \) component: \( r_y(t) = \int (-e^{-t} + 2) \, dt = e^{-t} + 2t + C_4 \). So, \( \mathbf{r}(t) = \left(\frac{t^2}{2} + 2t + C_3\right)\mathbf{i} + (e^{-t} + 2t + C_4)\mathbf{j} \).
05

Apply initial condition to find integration constants

We know \( \mathbf{r}(0) = \mathbf{i} - \mathbf{j} \). Setting \( t = 0 \) in \( \mathbf{r}(t) \) results in \( \mathbf{r}(0) = \left(\frac{0^2}{2} + 2(0) + C_3\right)\mathbf{i} + (e^0 + 2(0) + C_4)\mathbf{j} = C_3\mathbf{i} + (1 + C_4)\mathbf{j} \). This implies \( C_3 = 1 \) and \( 1 + C_4 = -1 \), resulting in \( C_4 = -2 \).
06

Construct the position vector

Substitute the constants \( C_3 = 1 \) and \( C_4 = -2 \) into the expression for \( \mathbf{r}(t) \). This results in \( \mathbf{r}(t) = \left(\frac{t^2}{2} + 2t + 1\right)\mathbf{i} + (e^{-t} + 2t - 2)\mathbf{j} \).

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Key Concepts

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

Acceleration Vector Integration
The process of finding velocity vectors from acceleration vectors involves integrating the components of the acceleration vector. Given an acceleration vector \( \mathbf{a}(t) = \mathbf{i} + e^{-t} \mathbf{j} \), you need to determine the velocity vector \( \mathbf{v}(t) \). This is achieved by integrating each component of the acceleration vector separately.- Integrating the \( \mathbf{i} \) component, which is constant (1), leads to \( v_x(t) = \int 1 \, dt = t + C_1 \), representing motion in the x-direction.- For the \( \mathbf{j} \) component, \( v_y(t) = \int e^{-t} \, dt = -e^{-t} + C_2 \), which describes motion in the y-direction.Once integrated, these components combine to form the velocity vector \( \mathbf{v}(t) = (t + C_1)\mathbf{i} + (-e^{-t} + C_2)\mathbf{j} \). This method of integration is fundamental to understanding how to translate acceleration into velocity, a crucial step in solving any problem involving motion dynamics.
Initial Conditions
Initial conditions provide the specific values that allow us to find unique solutions to differential equations. In this exercise, two initial conditions are given: \( \mathbf{v}(0) = 2\mathbf{i} + \mathbf{j} \) and \( \mathbf{r}(0) = \mathbf{i} - \mathbf{j} \).- For the velocity vector, substituting \( t = 0 \) gives \( \mathbf{v}(0) = C_1\mathbf{i} + (C_2 - 1)\mathbf{j} \). Using the initial condition \( 2\mathbf{i} + \mathbf{j} \), we can solve for the constants, yielding \( C_1 = 2 \) and \( C_2 = 2 \).- Similarly, for the position vector, substituting \( t = 0 \) in \( \mathbf{r}(t) \), we obtain \( C_3\mathbf{i} + (1 + C_4)\mathbf{j} \). Matching it with the given initial condition \( \mathbf{i} - \mathbf{j} \), results in \( C_3 = 1 \) and \( C_4 = -2 \).These initial conditions ensure that both the velocity and position vectors accurately reflect the motion of the particle from its starting configuration.
Integration Constants
During integration, constants of integration (e.g., \( C_1, C_2, C_3, \) and \( C_4 \)) arise. These constants reflect unknown factors that can be determined using initial conditions. They essentially allow the integration process to translate general solutions into specific solutions that fit the unique situation described in a problem.- For the velocity, the integration of \( a_x(t) \) and \( a_y(t) \) results in constants \( C_1 \) and \( C_2 \), which are placeholders until initial conditions provide their values.- Similarly, the integration to find the position vector introduces other constants \( C_3 \) and \( C_4 \).Once initial conditions are applied, these constants are no longer arbitrary and serve to provide a complete, individualized solution for the motion of the particle. Understanding and correctly identifying these constants are vital for accurately applying calculus to problems in physics and engineering, particularly in mechanics, where motion analysis is key.

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Most popular questions from this chapter

At time \(t=0\) a baseball that is \(5 \mathrm{ft}\) above the ground is hit with a bat. The ball leaves the bat with a speed of \(80 \mathrm{ft} / \mathrm{s}\) at an angle of \(30^{\circ}\) above the horizontal. (a) How long will it take for the baseball to hit the ground? Express your answer to the nearest hundredth of a second. (b) Use the result in part (a) to find the horizontal distance traveled by the ball. Express your answer to the nearest tenth of a foot.

A rock is thrown downward from the top of a building, 168 ft high, at an angle of \(60^{\circ}\) with the horizontal. How far from the base of the building will the rock land if its initial speed is \(80 \mathrm{ft} / \mathrm{s} ?\)

Determine whether the statement is true or false. Explain your answer. The velocity and unit tangent vectors for a moving particle are parallel.

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Find the displacement and the distance traveled over the indicated time interval. $$ \mathbf{r}=t^{2} \mathbf{i}+\frac{1}{3} t^{3} \mathbf{j} ; \quad 1 \leq t \leq 3 $$

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