Answer to Question #95609 in Mechanics | Relativity for Omar Bari

Question

A man runs 500m at 37° North by east then runs another 1000m at 60°west by north. Determine the module of the run and the orientation.

Answer 1

First a diagram of the situation is made.

The components are obtained for each displacement.

# Displacement 1

Horizontal axis component

$cos(37^{0})=\frac{D_{1x}}{D_{1}} \\D_{1x}=D_{1}cos(37^{0}) \\ D_{1x}=500m*cos(37^{0}) \\D_{1x}=399m$

Vertical axis component

$sin(37^{0})=\frac{D_{1y}}{D_{1}} \\D_{1y}=D_{1}sin(37^{0}) \\ D_{1y}=500m*sin(37^{0}) \\D_{1y}=301m$

Expressing as a vector

$\vec{D_{1}}=(399i+301j)[m]$

# Displacement 2

Horizontal axis component

$sin(60^{0})=\frac{D_{2x}}{D_{2}} \\D_{2x}=D_{2}sin(60^{0}) \\ D_{2x}=1000m*sin(60^{0}) \\D_{2x}=866m$

Vertical axis component

$cos(60^{0})=\frac{D_{2y}}{D_{2}} \\D_{2y}=D_{2}cos(60^{0}) \\ D_{2y}=1000m*cos(60^{0}) \\D_{2y}=500m$

Expressing as a vector

$\vec{D_{2}}=(-866i+500j)[m]$

The total displacement is

$\vec{}$ $\vec{D_{R}}=\vec{D_{1}}+\vec{D_{2}} \\ \vec{D_{R}}=(399i+301j)[m] +(-866i+500j)[m]\\ \vec{D_{R}}=[(399-866)i+(301+500)j][m] \\ \vec{D_{R}}=(-467i+801)[m]$

The run module is calculated using the Pythagorean theorem.

$|\vec{D_{R}}|=\sqrt{(-467)^{2}+(801)^{2}} \\ |\vec{D_{R}}|=927m$

the module of the run is $\boxed{|\vec{D_{R}}|=927m}$

The direction with respect to the + i axis is

$tan\theta=(\frac{801}{-467}) \\ \theta=tan^{-1}(\frac{801}{-467}) \\ \theta=-59.75^{0}+180^{0}=120^{0}$ Measured from the East direction (+ i)

Note: Being a negative component, the resulting angle will be reflected as in quadranta IV, but 180 degrees must be added to reflect it in quadrant II

with respect to the north the angle is $120^{0}-90^{0}=30^{0}$

Expressing the closest coordinate axis.

30°west by north

the the orientation of the run is $\boxed{ \text{ 30°west by north}}$