Find the component form of the vector v with
magnitude 12 and direction angle 300°.
A.
B.
C.
D.
Find the direction angle of p = –1, 4 to the nearest
tenth of a degree.
A. 14.5°
B. 76.3°
C. 104.5°
D. 166.7°
Applied Vector Operations
SOCCER A soccer player running forward at 7
meters per second kicks a soccer ball with a
velocity of 30 meters per second at an angle of 10°
with the horizontal. What is the resultant speed
and direction of the kick?
Since the soccer player moves straight forward, the
component form of his velocity v1 is 7, 0. Use the
magnitude and direction of the soccer ball’s velocity v2
to write this vector in component form.
Applied Vector Operations
v 2 = | v2 | cos θ, | v2 | sin θ
Component form of v2
= 30 cos 10°, 30 sin 10°
|v2| = 30 and θ = 10°
≈ 29.5, 5.2
Simplify.
Add the algebraic vectors representing v1 and v2 to
find the resultant velocity, r.
r = v1 + v2
Resultant vector
= 7, 0 + 29.5, 5.2
Substitution
= 36.5, 5.2
Vector Addition
Applied Vector Operations
The magnitude of the resultant is |r| =
or about 36.9. Next find the resultant direction θ.
Applied Vector Operations
a, b = 36.5, 5.2
Therefore, the resultant velocity of the kick is about
36.9 meters per second at an angle of about 8.1° with
the horizontal.
Answer: 36.9 m/s; 8.1°
SOCCER A soccer player running forward at 6
meters per second kicks a soccer ball with a
velocity of 25 meters per second at an angle of 15°
with the horizontal. What is the resultant speed
and direction of the kick?
A. 25.0 m/s; 15.1°
B. 25.0 m/s; 8.1°
C. 30.8 m/s; 15.1°
D. 30.8 m/s; 12.1°