CSULA- PHYSICS 211 Summer’12 Solution: Material to Study Chapter 3
OQ-3 (7e- Q3)
The vector 2D1 will be twice as long as D1 and in the opposite direction, namely northeast. Adding D 2 , which is about equally long
and southwest, we get a sum that is still longer and due east, choice (a).
8m/s, choice (c).
Take the difference of the coordinates of the ends of the vector. Final first means head end first: 1 (2) = 3 cm, answer (a).
Vectors A and B are perpendicular to each other.
Addition of a vector to a scalar is not defined. Think of numbers of apples and of clouds.
No, the magnitude of a vector is always positive. A minus sign in a vector only indicates direction, not magnitude.
We have r x 2 y 2 and tan 1 .
(a) The radius for this new point is: x 2 y 2 x 2 y 2 r and its angle is tan 1 180 .
(b) (2x)2 (2y)2 2r This point is in the third quadrant if x , y is in the first quadrant or in the fourth quadrant if x , y
is in the second quadrant. It is at an angle of 180 .
(c) (3x)2 (3y)2 3r This point is in the fourth quadrant if x , y is in the first quadrant or in the third quadrant
if x , y is in the second quadrant. It is at an angle of .
x 100 m tan 35.0 70.0 m
A x 25.0 Ay
A y 40.0 We observe that: tan .
A Ax Ay
25.02 40.02 47.2 units
So: tan 1 tan tan 1 1.60 58.0 .
The diagram shows that the angle from the +x axis can be found by subtracting from 180°: 180 58 122 .
(a) A B 3ˆ 2ˆj ˆ 4ˆj
i i 2ˆ 6ˆ
i j (b) A B 3ˆ 2ˆj iˆ 4ˆj
(c) A B 22 62 6.32 (d) A B 42 22 4.47
A B tan 1 71.6 288 A B tan 1 26.6
F F1 F2
F 120cos 60.0 i 120sin 60.0 ˆ 80.0cos 75.0 i 80.0sin 75.0 ˆ
F 60.0i 104ˆ 20.7ˆ 77.3ˆ 39.3i 181ˆ N
ˆ j i j ˆ
j (b) F3 F 39.3iˆ 181ˆj N
F 39.32 1812 185 N
E 17.0 cm cos27.0i 17.0 cm sin 27.0ˆ
j E 15.1iˆ 7.72ˆj cm
F 17.0 cm sin 27.0i 17.0 cm cos27.0ˆ
j F 7.72ˆi 15.1ˆj cm Note that we do not need to
explicitly identify the angle with the positive x axis.
G 17.0 cm sin 27.0i 17.0 cm cos27.0ˆ
j G 7.72iˆ 15.1ˆj cm y
Let represent the angle between the directions of A and B . Since A and B have the same magnitudes, A , B , and R A B
form an isosceles triangle in which the angles are 180 , , and . The magnitude of R is then R 2A cos . This can be seen
2 2 2
from applying the law of cosines to the isosceles triangle and using the fact that B A .
Again, A , – B , and D A B form an isosceles triangle with apex angle . Applying the law of cosines and the identity
1 cos 2sin 2 gives the magnitude of D as
D 2A sin . The problem requires that R 100D .
Thus, 2A cos 200A sin . This gives tan 0.010 and 1.15 .