An airplane flies 200 km due west from city A to city B and then 345 km in the direction of 32.0° north of west from city B to city C In straight-line distance, how far is city C from city A?

Add the two displacement vectors for the answer.

545

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

First, we need to break down the two legs of the plane's journey into their east-west (x) and north-south (y) components.

For the first leg, from City A to City B, the plane flies 200 km due west. Since it is a straight line, there is no north or south component, so the x-component is -200 km (negative because it's in the west direction), and the y-component is 0 km.

For the second leg, from City B to City C, the plane flies 345 km in the direction of 32.0° north of west. To find the x and y components, we need to use trigonometry.

The x-component can be found by taking the cosine of the angle and multiplying it by the magnitude (345 km):
x = 345 km * cos(32.0°) ≈ 290.36 km (rounded to two decimal places)

The y-component can be found by taking the sine of the angle and multiplying it by the magnitude (345 km):
y = 345 km * sin(32.0°) ≈ 181.18 km (rounded to two decimal places)

Now that we have the x and y components for both legs, we can add them up to find the final x and y components for the overall journey.

x_total = -200 km + 290.36 km ≈ 90.36 km (rounded to two decimal places)
y_total = 0 km + 181.18 km ≈ 181.18 km (rounded to two decimal places)

Using these final x and y components, we can calculate the straight-line distance (D) between City A and City C using the Pythagorean theorem:

D = sqrt(x_total^2 + y_total^2)
= sqrt((90.36 km)^2 + (181.18 km)^2)
≈ sqrt(8182.3296 km^2 + 32850.9924 km^2)
≈ sqrt(41033.3219 km^2)
≈ 202.57 km (rounded to two decimal places)

Therefore, City C is approximately 202.57 km away from City A in a straight-line distance.