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

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Chapter1: Units, Trigonometry. And Vectors
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### 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**

The
Transcribed Image Text:### 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** The
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