Question

A car travels 22 km due south and then 28 km in a direction 60° east of south. Find the Direction of the car's resultant vector. PLEASE HELP

Answer 1

A car travels 22 km due south and then 28 km in a direction 60° east of south. Find the Direction of the car's resultant Remember that any can be written as a coordinate like this: (AcosΘ,AsinΘ) where A is the magnitude of the object and Θ is the directional angle. So for the car travelling 20km north, that vector can be written as:

(20cos90°, 20sin90°)

Then when the car drives northwest at 60 degrees, that vector can be written as:

(35cos150°, 35sin150°) (do you see why I it's 150°?)

Then, simply calculate each of these coordinates, add them up, and then that resulting coordinate is the coordinate of the resulting displacement vector.

(20cos90°, 20sin90°) = (0, 20)

(35cos150°, 35sin150°) = (-30.31, 17.50)

Displacement vector (x, y) = (-30.31, 37.50)

With this, we can calculate the magnitude which is \displaystyle \sqrt{x^2+y^2}

x

2

+y

2

​

. Thus:

Magnitude = \displaystyle \sqrt{(-30.31)^2+(37.50)^2}=48.2km

(−30.31)

2

+(37.50)

2

​

=48.2km

The resulting directional angle is then \displaystyle \tan^{-1}(\frac{y}{x})+\pitan

−1

(

x

y

​

)+π. Adding \displaystyle \piπ is only necessary when x < 0 in order to put us in the proper quadrant. Thus:

Direction = \displaystyle \tan^{-1}(\frac{37.50}{-30.31})+\pi=2.25\ radians=128.9°tan

−1

(

−30.31

37.50

​

)+π=2.25 radians=128.9°

ges ok

HOPE THIS HELPS :):0