part d: vector addition using components\n1. calculate the total displacement in the horizontal (x) and…

part d: vector addition using components\n1. calculate the total displacement in the horizontal (x) and vertical (y) directions using the algebraic method and grasp. show all work in table 4.\ntable 4.\n| | δd_tx | δd_ty |\n|--|--|--|--|--|--|\n| g: | δd_ab = 11.4 m e | g: δd_bc = 8.56 m s |\n| | d_co = 27.6 m e | δd_de = 26.3 m s |\n| r: | δd_tx =? | r: δd_ty =? |\n| a: | δd_tx = δd_ab + δd_co | a: δd_ty = δd_bc + δd_de |\n| s: | δd_tx = δd_ab + δd_co | s: δd_ty = δd_bc + δd_de |\n| | δd_tx = 11.4 m e + 27.6 m e | δd_ty = 8.56 m s + 26.3 s |\n| | δd_tx = 39.0 m e | δd_ty = 34.86 m s |\n| | | δd_ty ≈ 34.9 m s |\n| p: | the total horizontal displacement is 39.0 m e | p: the total vertical displacement is 34.9 m s |

part d: vector addition using components\n1. calculate the total displacement in the horizontal (x) and vertical (y) directions using the algebraic method and grasp. show all work in table 4.\ntable 4.\n| | δd_tx | δd_ty |\n|--|--|--|--|--|--|\n| g: | δd_ab = 11.4 m e | g: δd_bc = 8.56 m s |\n| | d_co = 27.6 m e | δd_de = 26.3 m s |\n| r: | δd_tx =? | r: δd_ty =? |\n| a: | δd_tx = δd_ab + δd_co | a: δd_ty = δd_bc + δd_de |\n| s: | δd_tx = δd_ab + δd_co | s: δd_ty = δd_bc + δd_de |\n| | δd_tx = 11.4 m e + 27.6 m e | δd_ty = 8.56 m s + 26.3 s |\n| | δd_tx = 39.0 m e | δd_ty = 34.86 m s |\n| | | δd_ty ≈ 34.9 m s |\n| p: | the total horizontal displacement is 39.0 m e | p: the total vertical displacement is 34.9 m s |

Answer

Explanation:

Step1: Identify horizontal displacements

We have $\Delta\vec{d}{AB}=11.4m[E]$ and $\Delta\vec{d}{CD}=27.6m[E]$. The formula for total horizontal displacement $\Delta\vec{d}{Tx}$ is $\Delta\vec{d}{Tx}=\Delta\vec{d}{AB}+\Delta\vec{d}{CD}$.

Step2: Calculate horizontal displacement

$\Delta\vec{d}_{Tx}=11.4m + 27.6m=39.0m[E]$.

Step3: Identify vertical displacements

We have $\Delta\vec{d}{BC}=8.56m[S]$ and $\Delta\vec{d}{DE}=26.3m[S]$. The formula for total vertical displacement $\Delta\vec{d}{Ty}$ is $\Delta\vec{d}{Ty}=\Delta\vec{d}{BC}+\Delta\vec{d}{DE}$.

Step4: Calculate vertical displacement

$\Delta\vec{d}_{Ty}=8.56m+26.3m = 34.86m\approx34.9m[S]$.

Answer:

The total horizontal displacement is $39.0m[E]$ and the total vertical displacement is $34.9m[S]$.