An airplane flies 200 km due west from city A to city B and then 340 km in the direction of 35.0° north of west from city B to city C.

(a) In straight-line distance, how far is city C from city A?

(b) Relative to city A, in what direction is city C?
° north of west

In case you did not see the response:

physics - bobpursley, Tuesday, September 21, 2010 at 6:45pm

Break up the vectors into N, W components, and add, then use the results to reform the vector.

\frac{1}{2}

To find the straight-line distance between city A and city C, we can use the concept of vector addition.

First, let's understand the components of the airplane's motion:

From city A to city B:
Distance = 200 km
Direction = due west (0°)

From city B to city C:
Distance = 340 km
Direction = 35.0° north of west

(a) To find the straight-line distance, we can find the resultant displacement by adding the individual displacements (vectors) of the two legs.

Using trigonometry, we can break down the second leg into its northward (N) and westward (W) components.

Northward component = 340 km * sin(35°)
Westward component = 340 km * cos(35°)

Now, let's add the components:

Total northward displacement = Northward displacement from the second leg - No displacement from the first leg (as it was in the west direction)
Total westward displacement = Westward displacement from the second leg + Westward displacement from the first leg

Total northward displacement = 340 km * sin(35°)
Total westward displacement = 340 km * cos(35°) + 200 km

Using the Pythagorean theorem:

Straight-line distance (d) = sqrt((Total Northward Displacement)^2 + (Total Westward Displacement)^2)
d = sqrt[((340 km * sin(35°))^2) + ((340 km * cos(35°) + 200 km)^2)]

Calculate this expression to find the answer for (a).

(b) To find the direction of city C relative to city A, we can use the concept of trigonometry.

Direction = arctan(Total Northward Displacement / Total Westward Displacement)

Calculate this expression to find the answer for (b).