1 A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of d₁ = 157 cm and makes an angle of ₁ = 125° with the positive x axis. The resultant displacement has a magnitude of R = 138 cm and is directed at an angle of ₂ = 33.0° to the positive x axis. Find the magnitude and direction of the second displacement. Step 1 The diagram suggests that ď₂ is larger than either of the other vectors and is in a fourth-quadrant direction. y Ꮎ . Step 2 We use the component meth subtraction. a precise answer. We already know the Step 3 The resultant displacement of the mop is R = ₂ + ď₂ 1 Then d₂ = R-d₁ Substituting, ₂ = (138✔ 138 cm at 33.0°) - (157✔ ant displacement, so the algebra of solving a vector equation will guide us to 157 cm at 125°). a

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Chapter1: Units, Trigonometry. And Vectors
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A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of \( \vec{d}_1 = 157 \, \text{cm} \) and makes an angle of \( \phi_1 = 125^\circ \) with the positive x-axis. The resultant displacement has a magnitude of \( \vec{R} = 138 \, \text{cm} \) and is directed at an angle of \( \phi_2 = 33.0^\circ \) to the positive x-axis. Find the magnitude and direction of the second displacement.

**Step 1**

The diagram suggests that \( \vec{d}_2 \) is larger than either of the other vectors and is in a fourth-quadrant direction. 

*The diagram shows a vector triangle with vectors \( \vec{d}_1 \), \( \vec{d}_2 \), and \( \vec{R} \) arranged head-to-tail. The vector \( \vec{d}_2 \) points towards the fourth quadrant.*

**Step 2**

We will use the component method for a precise answer. We already know the resultant displacement, so the algebra of solving a vector equation will guide us to do a subtraction.

**Step 3**

The resultant displacement of the mop is \( \vec{R} = \vec{d}_1 + \vec{d}_2 \).

Then \( \vec{d}_2 = \vec{R} - \vec{d}_1 \).

Substituting, \( \vec{d}_2 = \langle 138 \cos 33^\circ, 138 \sin 33^\circ \rangle - \langle 157 \cos 125^\circ, 157 \sin 125^\circ \rangle \).

**Step 4**

The meaning of the negative sign reverses the direction of the second vector so we can rewrite the previous equation as 

\( \vec{d}_2 = (138 \text{ cm at } 33.0^\circ) + (157 \text{ cm at } 33^\circ) \).

The response you submitted has the wrong sign.

In components,

\[ 
\vec{d}_2 = [138 \cos 33^\circ \, \hat{i} + 138 \sin 125^\circ \, \hat{j}] - [157 \
Transcribed Image Text:A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of \( \vec{d}_1 = 157 \, \text{cm} \) and makes an angle of \( \phi_1 = 125^\circ \) with the positive x-axis. The resultant displacement has a magnitude of \( \vec{R} = 138 \, \text{cm} \) and is directed at an angle of \( \phi_2 = 33.0^\circ \) to the positive x-axis. Find the magnitude and direction of the second displacement. **Step 1** The diagram suggests that \( \vec{d}_2 \) is larger than either of the other vectors and is in a fourth-quadrant direction. *The diagram shows a vector triangle with vectors \( \vec{d}_1 \), \( \vec{d}_2 \), and \( \vec{R} \) arranged head-to-tail. The vector \( \vec{d}_2 \) points towards the fourth quadrant.* **Step 2** We will use the component method for a precise answer. We already know the resultant displacement, so the algebra of solving a vector equation will guide us to do a subtraction. **Step 3** The resultant displacement of the mop is \( \vec{R} = \vec{d}_1 + \vec{d}_2 \). Then \( \vec{d}_2 = \vec{R} - \vec{d}_1 \). Substituting, \( \vec{d}_2 = \langle 138 \cos 33^\circ, 138 \sin 33^\circ \rangle - \langle 157 \cos 125^\circ, 157 \sin 125^\circ \rangle \). **Step 4** The meaning of the negative sign reverses the direction of the second vector so we can rewrite the previous equation as \( \vec{d}_2 = (138 \text{ cm at } 33.0^\circ) + (157 \text{ cm at } 33^\circ) \). The response you submitted has the wrong sign. In components, \[ \vec{d}_2 = [138 \cos 33^\circ \, \hat{i} + 138 \sin 125^\circ \, \hat{j}] - [157 \
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