A car leaves it's garage and travels for 30km on a bearing of 145°,then 10kn on a bearing of 50°. Calculate:

Question

1. The distance travelled from the starting point
2. The bearing of the starting point from the car

Answer 1

To calculate the distance travelled from the starting point, we need to find the resultant vector of the two displacements.

1. The first displacement is 30km on a bearing of 145°, which can be represented as a vector with components x = 30cos(145°) and y = 30sin(145°).

x = 30cos(145°) ≈ -12.27 km
y = 30sin(145°) ≈ 25.06 km

2. The second displacement is 10km on a bearing of 50°, which can be represented as a vector with components x = 10cos(50°) and y = 10sin(50°).

x = 10cos(50°) ≈ 6.44 km
y = 10sin(50°) ≈ 7.66 km

Resultant x = -12.27 + 6.44 ≈ -5.83 km
Resultant y = 25.06 + 7.66 ≈ 32.72 km

Distance from the starting point = √((-5.83)^2 + (32.72)^2) ≈ 33.18 km

2. To find the bearing of the starting point from the car, we need to find the angle made by the resultant vector with the positive x-axis.

Bearing = arctan(32.72 / -5.83) ≈ -80.08°

The bearing of the starting point from the car is 80.08° clockwise from the direction the car is facing.