Solution
David answered on
Dec 29 2021
Solutions
1. We have the (r, θ) forms for points from which Cartesian (x, y) coordinate form can be derived as (rcosθ, rsinθ). So, we
have as follows.
• Point a:, r = 7.6Km, θ = 14◦ East of North ⇒ 90 − 14 = 76◦ from Horizontal⇒ Cartesian cordinates are
(7.6cos76, 7.6sin76) = (1.8386, 7.3742) = (x1, y1)
• Point b:, r = 5.1Km, θ = 27◦ west of South ⇒ −(90 + 27) = 117◦ from Horizontal⇒ Cartesian cordinates are
(5.1cos(−117), 5.1sin(−117)) = (−2.3154,−4.5441) = (x2, y2)
(a) Diagram is attached below.
Point a
Point
7.6km
14◦
27◦
0◦
x
y
5.1km
Figure 1: Scale Diagram of Points A & B
(b) Distance
w the points is given by Cartesian distance d =
√
(y2 − y1)2 + (x2 − x1)2 = 12.6215◦
(c) Relative to North(Y axis), a realtive to b has heading
a = y1 − y2 = 11.9183km
2. The arm moves from ra : (1, 2, 1)m to
: (3, 3, 2)m with force F : (60,−10, 30)N .
(a) Displacement vector from a to b is rab =
− ra = (2, 1, 1)m or 2̂i+ 1ĵ + 1k̂
(b) Work done by arm using force F for displacement rab is W = F.rab = (60̂i− 10ĵ + 30k̂).(2̂i+ 1ĵ + 1k̂) = 140J
3. Firework explodes into 4 fragments. Velocities of fragments are va = (2, 4,−1)m/s, vb = (3,−2,−2)m/s, vc = (3,−3,−4)m/s
& vd = 2a = (4, 8,−2)m/s.
(a) Magnitude of velocity of fragment d , |vd| =
√
42 + 82 + (−2)2 =...