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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
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Oct 20, 2022
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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
Mathematics
high-school
Forseti
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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
Filled Stacks
answered
Oct 24, 2022
by
Filled Stacks
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