2-D Motion HW #1
P3.6. Prepare: The position vector d whose magnitude d is 10 m has an x-component of 6 m. It makes an angle  with
the +x-axis in the first quadrant. We will use trigonometric relations to find the y-component of the position vector.
Solve: Using trigonometry, dx = d cos  , or 6 m = (10 m) cos  . This gives   53.1 . Thus the y-component of the
position vector d is dy = d sin   (10 m) sin 53.1 = 8 m.
Assess: The y-component is positive since the position vector is in the first quadrant.
P3.8. Prepare: The figure below shows the components vx and vy, and the angle . We will use Tactics Box 3.3 to find
the sign attached to the components of a vector.
Solve: (a) Since vx  v cos  , we have 2.5 m/s  (3.0 m/s) cos     cos 1 (2.5 m/s 30 m/s)  33.6   34 .
(b) The vertical component is v y  v sin   (3.0 m/s) sin 33.6  1.7 m/s.