An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) while returning to base camp. He was supposed to travel due north for 4.9 km, but when the snow clears, he discovers that he actually traveled 7.8 km at 50° north of due east.
(a) How far must he now travel to reach base camp?
(b) In what direction must he travel?
break the vectors down to components:
traveled: 7.8sin50 N + 7.8cos50E
Need to end up: 4.9N
Traveled+second leg=need to end up
7.8sin50N+7.8cos50E+SL=4.9N
SLeg=N(4.9-7.8sin50)- E7.8cos50
remember a negative direction is the opposite direction, as in - 5E=5W
To solve this problem, we can break down the explorer's displacement into its components using trigonometry. Let's label the sides of the triangle formed by the displacement as follows:
1. The side opposite to the 50° angle will represent the distance traveled north of due east.
2. The side adjacent to the 50° angle will represent the distance traveled due east.
3. The hypotenuse of the triangle will represent the total displacement.
(a) To find the distance the explorer must now travel to reach base camp, we need to determine the side opposite to the 50° angle. We can use the trigonometric function sine (sin) to calculate it.
Sine of an angle θ = Opposite side / Hypotenuse
In this case, the sine of 50° = Opposite side / 7.8 km.
So, Opposite side = sin(50°) * 7.8 km.
(b) To determine the direction the explorer must travel, we need to find the angle between his actual displacement and due north.
We can use the trigonometric function tangent (tan) to calculate it.
Tangent of an angle θ = Opposite side / Adjacent side
In this case, the tangent of the angle between the explorer's displacement and due north = Opposite side / 4.9 km.
So, the angle θ = tan^(-1)(Opposite side / 4.9 km).
To summarize:
(a) The distance the explorer must now travel to reach base camp is sin(50°) * 7.8 km.
(b) The direction the explorer must travel is the angle θ = tan^(-1)(Opposite side / 4.9 km).