Two beetles run across flat sand, starting at the same point. Beetle 1 runs 0.58 m due east, then 0.89 m at 32o north of due east. Beetle 2 also makes two runs; the first is 1.37 m at 35o east of due north. What must be (a) the magnitude and (b) the direction (relative to the east direction in the range of (-180°, 180°)) of its second run if it is to end up at the new location of beetle 1?

This is a displacement exercise, one of the easiest methods to solve it is to decompose the displacements in a coordinate system. Let's start with beetle 1

Let's use trigonometry to break down your second displacement

d₂ = 0.89 m θ = 32 north east

sin θ = [tex]d_{2y}[/tex] / d₂

d_{2y} = d2 sin 32

d_{2y} = 0.89 sin 32

d_{2y} = 0.472 m

cos 32 = d₂ₓ / d₂

d₂ₓ = d₂ cos 32

d₂ₓ = 0.89 cos 32

d₂ₓ = 0.755 m

We found the total displacement of the beetle 1

X axis

d₁ = 0.58 i ^

Dₓ = d₁ + d₂ₓ

Dₓ = 0.58 + 0.755

Dₓ = 1,335 m

Axis y

D_{y} = d_{2y}

D_{y} = 0.472 m

Now let's analyze the second beetle

d₃ = 1.37 m θ = 35 north east

Sin (90-35) = d_{3y} / d₃

d_{3y} = d₃ sin 55

d_{3y} = 1.35 sin 55

d_{3y} = 1,106 m

cos 55 = d₃ₓ / d₃

d₃ₓ = d₃ cos 55

d₃ₓ = 1.35 cos 55

d₃ₓ = 0.774 m

They ask us what the second displacement should be to have the same location as the beetle 1

Dₓ = d₃ₓ + dx’

D_{y} = d_{3y} + dy’

dx’= Dₓ - d₃ₓ

dx’= 1.335 - 0.774

dx’= 0.561 m

dy’= D_{y} - d_{3y}

dy’= 0.472 - 1,106

dy’= -0.634 m

We can give the result in two ways

d’= (0.561 i ^ - 0.634 j ^) m

Or in the form of module and address

d’= √ (dx’² + dy’²)

d’= √ (0.561² + 0.634²)

d’= 0.847 m

tan θ = dy’/ dx’

θ = tan⁻¹ dy ’/ dx’

θ = tan⁻¹ (-0.634 / 0.561)

θ = -48.5 º

This is 48.5 south east