Question

A hiker begins a trip by first walking 2500m southeast from her base camp. On the second day she walks 4000m in a direction 75.0° north of east, at which point she discovers a forest ranger's tower. Determine the magnitude and direction of the total displacement​

Answer 1

Answer:

 R = 37.7 i^i^ + 16.9 j^j^

We need to determine the components of the hiker’s displacement for each day.

We can denote the displacement on day 1 as A and day 2 as B.

On Day 1, Displacement A has a magnitude of 25 km and is directed 45∘∘ below the positive x axis.

⇒Ax=Acos(−θ)=25×cos(−45∘)=25×0.707=17.7km⇒Ax=Acos⁡(−θ)=25×cos⁡(−45∘)=25×0.707=17.7km

⇒Ay=Asin(−θ)=25×sin(−45∘)=25×−0.707=−17.7km⇒Ay=Asin⁡(−θ)=25×sin⁡(−45∘)=25×−0.707=−17.7km

On Day 2, Displacement B has a magnitude of 40 km and is directed 60∘∘ above the positive x axis.

⇒Bx=Bcos(θ)=40×cos(60∘)=40×0.5=20km⇒Bx=Bcos⁡(θ)=40×cos⁡(60∘)=40×0.5=20km

⇒By=Bsin(θ)=40×sin(60∘)=40×0.866=34.6km⇒By=Bsin⁡(θ)=40×sin⁡(60∘)=40×0.866=34.6km

Now, Rx=Ax+Bx=17.7+20=37.7Rx=Ax+Bx=17.7+20=37.7 and Ry=Ay+By=−17.7+34.6=16.9Ry=Ay+By=−17.7+34.6=16.9

⇒R=37.7i^+16.9j^⇒R=37.7i^+16.9j^