Question

A car travels 10km southeast and then 15 km in a direction 60 degrees north of east. What is the magnitude of the car's resultant vector?

Answer 1

Answer: 15.85km

Step-by-step explanation:

The car starts at the point (0,0)

Let's define east as the positive x-axis and north as the positive y-axis.

first, the car travels 10km southeast, so the angle is -45° from the east.

Then the new position of the car is:

(10km*cos(-45°), 10km*sin(-45°)) = (7.07km, -7.07km)

Now, from this point the car travels 15km with an angle of 60° (counting from the east)

So the new position is:

(7.07km + 15km*cos(60°), -7.07km + 15km*sin(60°)) = (14.57km  , 5.91km)

The magnitude of a vector (x, y) is √(x^2 + y^2)

So the magnitude of the vector (14.57km  , 5.91km) is:

√((14.57km)^2 + (5.91km)^2) = 15.85km