A car is driven east for a distance of 49 km, then north for 35 km, and then in a direction 33° east of north for 27 km. Draw the vector diagram and determine the total displacement of the car from its starting point.

What is the magnitude and direction?

We don't do graphs.

Add the three displacement vectors to get the total displacement.

I recommend adding the x (east) and y (north) components for the total displacement vector. You won't need a vector diagram to answer the problem (unless you really have to turn one in).

A CAR IS 150 KM FROM EAST TO NORTH THE MAGNITUDE IS 13.4

45.0m

To solve this problem, we can use vector addition to find the total displacement of the car. We need to break down each leg of the car's journey into its x and y components.

1. Eastward leg: The car travels 49 km east. Since this is a purely horizontal movement, the x-component of this leg is 49 km, and the y-component is 0 km.

2. Northward leg: The car travels 35 km north. Since this is a purely vertical movement, the x-component of this leg is 0 km, and the y-component is 35 km.

3. Direction 33° east of north leg: The car travels 27 km in a direction 33° east of north. To find the x and y components of this leg, we need to resolve the vector into its x and y components.

Let's call the angle between the north direction and the direction 33° east of north as angle A. We can find angle A by subtracting 33° from 90° (since 90° represents the north direction).

Angle A = 90° - 33° = 57°

Using trigonometry, we can find the x and y components:

x-component = 27 km * sin(A)
y-component = 27 km * cos(A)

x-component = 27 km * sin(57°)
x-component ≈ 23.3 km

y-component = 27 km * cos(57°)
y-component ≈ 17.8 km

Now, we have the x and y components for each leg of the journey:

Leg 1: x1 = 49 km, y1 = 0 km
Leg 2: x2 = 0 km, y2 = 35 km
Leg 3: x3 = 23.3 km, y3 = 17.8 km

To find the total displacement, we need to add these vectors together:

x-total = x1 + x2 + x3
x-total = 49 km + 0 km + 23.3 km
x-total ≈ 72.3 km

y-total = y1 + y2 + y3
y-total = 0 km + 35 km + 17.8 km
y-total ≈ 52.8 km

The total displacement (x-total, y-total) is approximately (72.3 km, 52.8 km). To find the magnitude and direction of the displacement, we can use the Pythagorean theorem and trigonometry.

Magnitude of displacement (magnitude) = sqrt((x-total)^2 + (y-total)^2)
magnitude = sqrt((72.3 km)^2 + (52.8 km)^2)
magnitude ≈ sqrt(5225.29 km^2 + 2787.84 km^2)
magnitude ≈ sqrt(8013.13 km^2)
magnitude ≈ 89.58 km

Direction of displacement (θ) = tan^(-1)(y-total / x-total)
θ = tan^(-1)(52.8 km / 72.3 km)
θ ≈ 37.1°

Therefore, the total displacement of the car from its starting point is approximately 89.58 km at an angle of 37.1° east of north.