HW Solutions 2  8.01 MIT  Prof. Kowalski
Topics: Vectors and Two dimensional motion
1)
1.38
Please refer to figure 1.26 p.35.
To add the several vector displacements, we use components, refer
ring to east and north instead of x and y. The net northward dis
placement is 2.6 + (3.1) (sin 45) =+ 4.8 km, and the net eastward
displacement is (4.0)+ (3.1) (cos 45) = + 6.2 km. The magnitude of
the net displacement is
(4
.
8)
2
+ (6
.
2)
2
= 7
.
8 km, using Pythago
ras’ theroem. To find the direction, note that the ratio of northward
to eastward displacement is the tangent of the angle with the east
axis: arctan(
4
.
8
6
.
2
) = 38 noth of east.
* The total displacement is a vector starting from the starting point
towards the end.
By sketching the vectors to scale you’ll see the
agreement with the method of components.
2)
1.70
Please refer to figure 1.32 p.37.
The displacement vectors of the three legs must sum up to give the
total observed displacement.
This vector equality provides TWO
equations, one for each component  the net x displacement should
be 5.80 km and the net y displacement should be 0:
The third leg must take the sailor east the distance: (5.80)  (3.50)
cos 45  (2.00) = 1.33 km and a distance north (3.5) sin 45 = 2.47 km.
⇒
magnitude of the displacement is
(1
.
33)
2
+ (2
.
47)
2
= 2
.
81 km
and the direction arctan(
2
.
47
1
.
33
) = 62north of east.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
3)
1.50
Please refer to figure 1.28 p.35.
Method : (Product of magnitudes
×
cos
θ
)
The angles must be found from the figure.
A.B
=
AB
cos
θ
AB
= (12
×
15) cos 93 =
−
9
.
4
m
2
B.C
=
BC
cos
θ
BC
= (15
×
6) cos 80 = +15
.
6
m
2
C.A
=
CA
cos
θ
CA
= (12
×
6) cos 173 =
−
71
.
5
m
2
4)
1.52
The problem as initially posed was ambiguous as to which figure
was referred to.” Either of these solutions will be graded as correct.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 WHITE
 Physics

Click to edit the document details