Chapter 3 P19 Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Serway, Faughn and Vuille: College Physics 8th Edition , Thomson Brooks/Cole, Vol I(ISBN #) 978-049511374-3
THE PROBLEM STATEMENT
Ch 3 P19 A commuter airplane starts from an airport and takes the route shown in Figure P3.19.
The plane first flies to city A, located 175 km away in a direction 30.0° north of east. Next, it
flies for 150 km 20.0° west of north, to city B. Finally, the plane flies 190 km due west, to city C.
Find the location of city C relative to the location of the starting
point.
Page 1 of 3
Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Ch 3 P19 A commuter airplane starts from an airport and takes the route shown in Figure P3.19.
The plane first flies to city A, located 175 km away in a direction 30.0° north of east. Next, it
flies for 150 km 20.0° west of north, to city B. Finally, the plane flies 190 km due west, to city C.
Find the location of city C relative to the location of the starting
point.
BRAINSTORMING-Definitions, concepts ,
principles and Discussion
The component of the Resultant Vector equals the sum
of that component of the added vectors;
Rx= (A+B+C+ )x = Ax + Bx + Cx + (1)
Ry = (A+B+C+ )y = Ay + By + Cy + (2)
Note: The red “+” at the end is called a
“continuation’ that say if you have more vectors
just continue adding their x or y components.
(3)
R = tan-1 (Ry/Rx)
by=bcos20o
Rx=acosR
ax=acos30o
o
The angle of the resultant vector is
R
ay=bsin30
R = sqrt(Rx2 + Ry2)
o
Ry=asinR
The magnitude of the resultant, R, is given by the
Pythagorean Theorem as
bx=bsin20
(4)
See the figure where the components are superposed onto the original picture.
Note: Equations (3) and (4) are only true if the coordinate axes are all perpendicular to each
other.
Page 2 of 3
Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Ch 3 P19 A commuter airplane starts from an airport and takes the route shown in Figure P3.19.
The plane first flies to city A, located 175 km away in a direction 30.0° north of east. Next, it
flies for 150 km 20.0° west of north, to city
B. Finally, the plane flies 190 km due west,
o
bx=bsin20
to city C.
Rx=acosR
ax=acos30o
o
Rx= (A+B+C+ )x = Ax + Bx + Cx +
R
ay=bsin30
Basic Solution
by=bcos20o
Find the location of city C relative to the
location of the starting
point.
Ry=asinR
Ry = (A+B+C+ )y = Ay + By + Cy +
The magnitude of the resultant, R, is given by the Pythagorean Theorem as
R = sqrt(Rx2 + Ry2)
The angle of the resultant vector is
R = tan-1 (Ry/Rx)
ax = acosx = 175km cos30o = 175km 0.866 = 152 km
ay = asiny = 175km sin30o = 175km 0.500 = 87.5 km
bx = bsinx = -150km sin20o = -150km 0.342 = -51.3 km
by = bcosy = 150km cos20o = 150km 0.940 = 141 km
cx = ccosx = 190km cos180o = 190km (-1) = - 190 km
cy = csiny = 190km sin180o = 0
Rx = ax + bx + cx = 152 - 51.3 – 190 km = -89.3. km
Ry = ay + by + cy = 87.5 + 141 + 0
= 229. km.
R = sqrt(Rx2 + Ry2) = sqrt(89.32 +2292) = sqrt(7974.5+52441)=sqrt(60415) 246 km.
R = tan-1 (Ry/Rx) = tan-1 (229km/-89.3km) = tan-1 (-2.564) = 90o -39o = 51o to –x axis.
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