Question

Bob walks 370 m south, then jogs 410 m southwest, then walks 370 m in a direction 28 degrees east of north.

Use either trigonometry or components to find the displacement that will return Bob to his starting point by the most direct route. Give your answer as a distance.

Find the direction of the displacement that will return Bob to his starting point by the most direct route.

Answer 1

We will measure all angles from West, the negative x-axis and divide the journey into 3 parts:
P1 = 370y
P2 = 410cos(45)x + 410sin(45)y = 290x + 290y
P3 = 370cos(270 - 28)x + 370sin(270 - 28) = -174x - 327y

Overall displacement:
x = 290 - 174 = 116 m
y = 370 + 290 - 327 = 333 m

displacement = √(116² + 333²)
= 353 m

Direction:
tan(∅) = y/x
∅ = tan⁻¹ (333 / 116)
∅ = 70.8° from West.

Answer 2

Consider east-west direction along x-axis with north being in positive x-direction.

Consider north-south direction along x-axis with north being in positive x-direction.

A = magnitude of displacement in south direction = 370 m

$A_{x}$ = x-component of A = 0 m

$A_{y}$ = y-component of A = - 370 m

B = magnitude of displacement in south-west direction = 410 m

$B_{x}$ = x-component of B = - 410 Cos45 = - 289.91 m

$B_{y}$ = y-component of B = - 410 Sin45 = - 289.91 m

C = magnitude of displacement in 28 deg east of north direction = 370 m

$C_{x}$ = x-component of C = 370 Sin28 = 173.7 m

$C_{y}$ = y-component of C = 370 Cos28 =  326.7 m

$D_{x}$ = x-component of D = ?

$D_{y}$ = y-component of D = ?

To return to starting position, the sum of individual displacements alng each axis must be zero.

$A_{x}$ + $B_{x}$ + $C_{x}$ + $D_{x}$ = 0

0 -  289.91 + 173.7 + $D_{x}$ = 0

$D_{x}$ = 116.21 m

$A_{y}$ + $B_{y}$ + $C_{y}$ + $D_{y}$ = 0

- 370 - 289.91 + 326.7 + $D_{y}$ = 0

$D_{y}$ = 333.21 m

magnitude of D is given using Pythagorean theorem as

magnitude = sqrt(($D_{x}$)² + ($D_{y}$)²)

magnitude = sqrt((116.21)² + (333.21)²) = 352.89 m

direction : tan⁻¹(333.21/116.21) = 70.8 deg