Problem 1:
A particle undergoes three consecutive displacements: 𝑑! = 15𝚤 + 30𝚥 + 12𝑘 𝑚, 𝑑! = 23𝚤 − 14𝚥 − 5𝑘 𝑚 and 𝑑! = −13𝚤 + 15𝚥 𝑚. Find
the components of the resultant displacement and its magnitude.
Ans: 𝑅 = 25𝚤 + 31𝚥 + 7𝑘 𝑚 and 𝑅 = 40𝑚
Problem 2:
The polar coordinates of a point are 𝑟 = 5.5𝑚 and 𝜃 = 240° . What are the Cartesian coordinates of this point?
Sln: 𝑥 = 𝑟𝑐𝑜𝑠𝜃 = 5.5𝑚 𝑐𝑜𝑠240° = 5.5𝑚 −0.5 = −2.75𝑚
𝑦 = 𝑟𝑠𝑖𝑛𝜃 = 5.5𝑚 sin 240° = 5.5𝑚 −0.866 = −4.76𝑚
Problem 3:
If 𝐴 = (3𝚤 − 4𝚥 + 4𝑘) ve 𝐵 = 2𝚤 + 4𝚥 − 8𝑘);
!
(a) Express 𝐷 in unit vector notation, 𝐷 = (2𝐴 − ! 𝐵).
(b) Find the magnitude and direction of 𝐷.
(-2.75,-4.76)m
Problem 4:
A fly lands on one wall of a room. The lower left-hand corner of the wall is selected as the origin of a two-dimensional Cartesian coordinate
system. If the fly is located at the point having coordinates 2, 1 𝑚,
a) How far is it from the corner of the room?
Sln: The x distance out to the fly is 2m and the y distance up to the fly is 1m. We can use the Pythagorean theorem to find the distance from the
origin to the fly,
𝑥 ! + 𝑦 ! = 2! + 1! = 5 = 2.24𝑚
Sln: 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =
b) What is its location in polar coordinates?
Sln: 𝜃 = 𝑡𝑎𝑛!!
!
= 22.6° ;
!
𝑟 = (2.24𝑚, 22.6° )
Problem 5:
!⃗ ! = 20𝑚 , and
The magnitudes of the vectors shown in the figure are !𝐴⃗! = 10𝑚 , !𝐵
!𝐶⃗! = 15𝑚.
a) Find the components of each vector.
30° b) Write each vector in unit vector notation.
20° !
!⃗ − 𝐶⃗! = 𝐷
!⃗ find 𝐷
!⃗ in unit vector notation.
c) If !2𝐴⃗ + ! 𝐵
Problem 6:
Two points in a plane have polar coordinates (2.5𝑚, 30° ) and (3.8𝑚, 120° ). Determine;
a) The Cartesian coordinates of these points.
Sln: 𝑥 = 𝑟𝑐𝑜𝑠𝜃 𝑎𝑛𝑑 𝑦 = 𝑟𝑠𝑖𝑛𝜃, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒
𝑥! = 2.5𝑚 𝑐𝑜𝑠30° , 𝑦! = 2.5𝑚 𝑠𝑖𝑛30° )
𝑥! , 𝑦! = (2.17, 1.25)𝑚
𝑥! = 3.8𝑚 𝑐𝑜𝑠120° , 𝑦! = 3.8𝑚 𝑠𝑖𝑛120° )
𝑥! , 𝑦! = (−1.9, 3.29)𝑚
b) The distance between them.
Sln: 𝑑 =
(∆𝑥)! + (∆𝑦)! =
(−1.9 − 2.17)! + (3.29 − 1.25)! = 4.55𝑚
Problem 7:
Vector A has x and y components of -8.70 cm and 15.0 cm, respectively; vector B has x and y components of 13.2 cm and -6.60 cm,
respectively. If A + B - 3C = 0, what are the components of C ?
Problem 8:
Vector 𝐴 has a negative x-component of 3 meters in length and positive y-component of 2 meters in length.
(a)
Determine an expression for 𝐴 in unit vector notation.
(b)
Determine the magnitude and direction of 𝐴.
(c)
What vector 𝐵 when added to 𝐴 gives a resultant vector with no x-component and a negative y-component 4 units in length.
Problem 9:
The polar coordinates of two vectors are; 𝐴 = (10𝑚, 240° ), and 𝐵 = (5𝑚, 180° ).
a) Express each vector in unit vector notation.
!
b) Find, 𝑅 = 2𝐴 + ! 𝐵 in unit vector notation.
c) Find the magnitude of 𝑅.
d) Find the anticlockwise angle that vector 𝑅 makes with +x axis.
Problem 10:
The polar coordinates of three vectors are; 𝐴 = (12𝑚, 150° ), 𝐵 = (5𝑚, 50° ), and 𝐶 = (20𝑚, 230° ). If vector 𝑅 is defined as, 𝑅 = 𝐴 + 2𝐵 −
𝐶, find the polar coordinates of vector 𝑅.
Problem 11:
A hiker begins a trip by first walking 25𝑘𝑚 southeast from her car. She stops and sets up her tent for the night. On the second day, she walks 40𝑘𝑚 in a
direction 60° north of east, at which point she discovers a forest ranger’s tower.
a) Determine the components of the hiker’s displacement for each day.
Ans: 𝐴! = 17.7𝑘𝑚, 𝐴! = −17.7𝑘𝑚, 𝐵! = 20𝑘𝑚, 𝐵! = 34.6𝑘𝑚
b) Determine the components of the hiker’s resultant displacement 𝑅 for the trip. Find an expression for 𝑅 in terms of unit vectors.
Ans: 𝑅 = 37.7𝚤 + 16.9𝚥 𝑘𝑚
Problem 12:
A commuter airplane takes the route shown in figure. First, it flies from origin of the coordinate system shown to city A, located 175𝑘𝑚 in a direction 30°
north of east. Next, it flies 153𝑘𝑚 20° west of north to city B. finally, it flies 195𝑘𝑚 due west to city C. find the location of city C relative to origin.
Ans:𝑅!⃗ = (−95.3𝚤̂ + 232𝚥̂)𝑘𝑚 so !𝑅!⃗ ! = 251𝑘𝑚 22.3° west of north)