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?

 d’= (0.561 i ^ - 0.634 j ^) m ,  d’= 0.847 m ,   48.5 south east

Explanation:

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 = / 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