A car is driven 105 km west and then 30 km southwest. What is the displacement of the car from the point of origin (magnitude and direction)?

Displacement is a vector. To find the displacement due to multiple legs of a trip, we sum the components (x,y) or (east,north) by resolving each leg into the respective components.

105 km west = (-105,0)
30 km SW = (30cos(225°),30sin(225°))
Sum each component to get the resultant components. The direction can be found with the tan-1(y/x) function.

Post your answer for a check if you wish.

7.58

-121.88

To find the displacement of the car from the point of origin, we can use vector addition.

First, let's break down the car's movement into two components: west and southwest.

The westward distance of 105 km can be represented as a vector with a magnitude of 105 km in the west direction (left on a map).

The southwest distance of 30 km can be represented as a vector with a magnitude of 30 km in the southwest direction (45 degrees between south and west, in the bottom left direction on a map).

Now, let's add these two vectors to find the resultant displacement of the car:

1. Convert the southwest vector into rectangular or Cartesian coordinates:
The southwest vector can be resolved into two components, one in the west direction and one in the south direction. Since it forms a 45-degree angle with the west direction, both components will have the same magnitude.
The westward component can be calculated using cosine: west_component = magnitude * cos(45)
The southward component can be calculated using sine: south_component = magnitude * sin(45)

In this case, the magnitude is 30 km, so:
west_component = 30 km * cos(45) ≈ 21.2 km
south_component = 30 km * sin(45) ≈ 21.2 km

2. Add the westward distance vector and the west_component vector:
west_displacement = 105 km + west_component ≈ 105 km + 21.2 km = 126.2 km

3. Add the southward distance vector and the south_component vector:
south_displacement = south_component ≈ 21.2 km

4. Combine the west and south displacements using the Pythagorean theorem:
displacement = √(west_displacement^2 + south_displacement^2) = √(126.2 km^2 + 21.2 km^2) ≈ √(17532.68 km^2) ≈ 132.42 km

The displacement of the car from the point of origin is approximately 132.42 km.

To describe the direction, we can use the inverse tangent:
direction = tan^(-1)(south_displacement / west_displacement)
direction = tan^(-1)(21.2 km / 126.2 km) ≈ tan^(-1)(0.168) ≈ 9.19° south of west

Therefore, the magnitude of the displacement is approximately 132.42 km, and the direction is 9.19° south of west.