ADDITIONAL SOLVED PROBLEMS
UNIT- 05 VECTORS
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(Note: In the following vectors are represented either with an arrow over the symbol or symbols in
bold type.)
Prob.1. Mark the following statements as true [T] or false [F] by circling your choice.
I. Consider two vectors
and
[a]
[b]
and
[c] Ax = Bx and Ay = By
.
[T]
[F]
[T]
[F]
[T]
[F]
if and only if:
Prob.2. The vector A has components Ax= 18.0 units and Ay= - 6.0 units. Vector B has components
Bx= - 3.0 units and By= - 7.0 units. Find the magnitude and direction of the vector C such that
A + 2B – 4C = 0
Solution:
A + 2B – 4C = 0 = (18 – 6 – 4 Cx)i + ( - 6 – 14 – 4 Cy) j = 0
or Cx = 3 units and Cy = - 5 units
The magnitude of C = [(3)2+ (- 5)2]1/2 = [34]1/2 = 5.8 units.
Orientation of C with respect to the x-axis,
θ = tan-1[ -5/3] = - 590.
Prob. 3. The vector A has components Ax= -36.0 units and Ay= 24.0 units. Vector B has components
Bx= 24.0 units and By= 36 units.
[a] Compute the magnitude of A
The magnitude of A = [(-36)2 + (24)2]1/2 = [1872]1/2 = 43.3 units.
[b] Compute the magnitude of B
The magnitude of B = [(24)2 + (36)2]1/2 = [1872]1/2 = 43.3 units.
[c] Compute the components of A+ B
(Ax + Bx) = - 36.0 +24.0 = - 12.0 units
(Ay + By) = 24.0 +36.0 = 60.0 units
[d] Compute the components of 2A - B
(2Ax - Bx) = -72.0 -24.0 = - 96.0 units
(2Ay – By) = 48.0 -36.0 = 12.0 units
Prob.4. A jogger J jogs from O to D in four straight line segments as shown in the diagram, where
OA = 50.0m, AB = BC = 100.0m, and CD = 70.0m. CD
is parallel to the y-axis.
[a] What is the displacement of the jogger as she travels
from O to D?
Solution:
    
OD = OA + AB + BC + CD
= 50i +100(cos53oi – sin53oj)
– 100(cos30oi + sin30oj)+70j
= (50+60 – 87)i + ( -80 – 50+70)j
= 23i – 60j
[b] What is the magnitude of the jogger’s displacement
as she travels from O to D?
The magnitude of the displacement
= [(23)2 +(60)2 ]1/2 = 64.2m
[c] What is the distance travelled by the jogger as she jogs from O to D?
Distance, d = OA + AB + BC + CD = 320 m.
Prob.5. A jogger J jogs from O to d in four straight line segments in 50.0 s as shown in the diagram,
where Oa = 80.0m, ab=150.0m, bc = 50.0m and cd = 100.0m.
[a] Express each displacement vector A, B, C, and D in the usual unit vector notation.
A = - 80.0j m
B = 150 (sin 30o i + cos30o j) = 75 i + 130 j m
C = - 50(sin65o i + cos65o j) = - 45.3 i – 21.1 j m
D = 100 (- sin53o i + cos53o j) = - 80 i + 60 j m
(Note: Parts [b], [c], and [d] below are optional. These parts contain concepts of speed and velocity that
are discussed in UNIT-06 )
[b]. Express the average velocity of J for the trip from O to d in the unit vector notation.
V = (A+B+C+D)/50.0 = [(75 – 45.3 – 80)i + ( - 80+130 - 21.1 +60)j]/50.0
= (- 50.3 i + 88.9j)/50.0 = - i + 1.78j m/s
[c]. Determine the magnitude and direction of the average velocity of J for the trip from O to d.
V = [1 + (1.78)2]1/2 = 2.0 m/s
θ = tan-1(1.78) = 60.7o
[d]. What is the average speed of J for the trip from O to d?
Average speed, s = distance/time = (80.0+150.0+50.0+100.0)/50.0 = 380.0/50.0 = 7.6 m/s