### Motion in the x-y Plane#### Problem StatementAn object moving in the x-y plane has a position vector given by: \[ \vec{r} = (2t^3 - 5t)\hat{i} + (6 - t^4)\hat{j} \](in meters).Find the particle’sa. Position vector at \( t = 0 \) secondsb. Position vector at \( t = 2 \) secondsc. Displacement vector for this change in position#### Solution**a. Position vector at \( t = 0 \) seconds**To find the position vector at \( t = 0 \) seconds, substitute \( t = 0 \) into the position vector equation:\[ \vec{r}(0) = (2(0)^3 - 5(0))\hat{i} + (6 - (0)^4)\hat{j} \]\[ \vec{r}(0) = 0\hat{i} + 6\hat{j} \]Thus, the position vector at \( t = 0 \) seconds is:\[ \vec{r}(0) = 6\hat{j} \, \text{meters} \]**b. Position vector at \( t = 2 \) seconds**To find the position vector at \( t = 2 \) seconds, substitute \( t = 2 \) into the position vector equation:\[ \vec{r}(2) = (2(2)^3 - 5(2))\hat{i} + (6 - (2)^4)\hat{j} \]\[ \vec{r}(2) = (2(8) - 5(2))\hat{i} + (6 - 16)\hat{j} \]\[ \vec{r}(2) = (16 - 10)\hat{i} + (6 - 16)\hat{j} \]\[ \vec{r}(2) = 6\hat{i} - 10\hat{j} \]Thus, the position vector at \( t = 2 \) seconds is:\[ \vec{r}(2) = 6\hat{i} - 10\hat{j} \, \text{meters} \]**c. Displacement vector for this change in position**The

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Description: ### Motion in the x-y Plane#### Problem StatementAn object moving in the x-y plane has a position vector given by: \[ \vec{r} = (2t^3 - 5t)\hat{i} + (6 - t^4)\hat{j} \](in meters).Find the particle’sa. Position vector at \( t = 0 \) secondsb. Positi

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5. An object moving in the x-y plane has a position vector given by: 7 = (2t³ - 5t)i + (6-t¹)ĵ (in meters). Find the particle's a. position vector at t = 0s b. position vector at t = 2s 4 c. displacement vector for this change in position

5. An object moving in the x-y plane has a position vector given by: 7 = (2t³ - 5t)i + (6-t¹)ĵ (in meters). Find the particle's a. position vector at t = 0s b. position vector at t = 2s 4 c. displacement vector for this change in position

College Physics

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Displacement, Velocity and Acceleration

In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.

Linear Displacement

The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.

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Question

### Motion in the x-y Plane

#### Problem Statement
An object moving in the x-y plane has a position vector given by:
\[ \vec{r} = (2t^3 - 5t)\hat{i} + (6 - t^4)\hat{j} \]
(in meters).

Find the particle’s

a. Position vector at \( t = 0 \) seconds

b. Position vector at \( t = 2 \) seconds

c. Displacement vector for this change in position

#### Solution

**a. Position vector at \( t = 0 \) seconds**

To find the position vector at \( t = 0 \) seconds, substitute \( t = 0 \) into the position vector equation:

\[ \vec{r}(0) = (2(0)^3 - 5(0))\hat{i} + (6 - (0)^4)\hat{j} \]

\[ \vec{r}(0) = 0\hat{i} + 6\hat{j} \]

Thus, the position vector at \( t = 0 \) seconds is:

\[ \vec{r}(0) = 6\hat{j} \, \text{meters} \]

**b. Position vector at \( t = 2 \) seconds**

To find the position vector at \( t = 2 \) seconds, substitute \( t = 2 \) into the position vector equation:

\[ \vec{r}(2) = (2(2)^3 - 5(2))\hat{i} + (6 - (2)^4)\hat{j} \]

\[ \vec{r}(2) = (2(8) - 5(2))\hat{i} + (6 - 16)\hat{j} \]

\[ \vec{r}(2) = (16 - 10)\hat{i} + (6 - 16)\hat{j} \]

\[ \vec{r}(2) = 6\hat{i} - 10\hat{j} \]

Thus, the position vector at \( t = 2 \) seconds is:

\[ \vec{r}(2) = 6\hat{i} - 10\hat{j} \, \text{meters} \]

**c. Displacement vector for this change in position**

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