Vector Addition And Subtraction Problems And Solutions

 Problem #1

Write vectors equations for each diagram below

Answer:
[a] C = –A + (–B) 
[b] C = A + B 
[c] C = –B + (–A) 
[d] C – A + D + B = 0 
[e] A + C = D + B 
[f] A – C = B + D

Problem #2
Sketch vector diagrams for the following vector equations
[a] A = C + D + B 
[b] A = C + B – D  
[c] A + B = C + D 
[d] A + B = C + D

Answer:

Problem #3
The given vector components correspond to the vector r as drawn at right.
(a) Use the inverse tangent function to find the distance angle q.
(b) Use the Pythagorean Theorem to determine the magnitude of r.

Answer:
(a) the distance angle q is
θ = tan^-1(r_y/r_x) = tan^-1(–9.5 m/14 m) = –34⁰ or 34° below the +x axis
(b) the magnitude of r is
r = (r_x² + r_y²)^1/2 = [(14 m)² + (-9.5 m)²] = 17 m

Problem #4
The two vectors A (length 50 units) and B (length 120 units) are drawn at right.

Answer
(a) Find B_x:
B_x =  (120 units) cos 70⁰ = 41 units
Since the vector A points entirely in the x direction, we can see that A_x = 50 units and that vector A has the greater x component.
(b) Find B_y:
B_y =  (120 units) sin 70⁰ = 113 units
The vector A has no y component, so it is clear that vector B has the greater y component. However, if one takes into account that the y-component of B is negative, then it follows that it smaller than zero, and hence A has the greater y-component.

Problems #5
Two vectors A (length 40.0 m) and B (length 75.0 m) are drawn to the right, as shown. Determine the resultant kedau vector and its direction.

Answer:
A sketch (not to scale) of the vectors and their sum is shown at right.
Add the x components: C_x = A_x + B_x = (40.0 m) cos (–20.0⁰) + (75.0 m) cos (50.0⁰) = 85.8 m
Add the y components: C_y = A_y + B_y = (40.0 m) sin (–20.0⁰) + (75.0 m) sin (50.0⁰) = 43.8 m
Find the magnitude of C: C = (C_x² + C_y²)^1/2 = [(85.8 m)² + (43.8 m)²]^1/2 = 96.3 m
Find the direction of C: θ = tan^-1(C_y/C_y) = tan^-1(43.8 m/85.8 m) = 27.0⁰

Problem #6
The vector A has components A_x = 18.0 units and A_y = - 6.0 units. Vector B has components B_x= - 3.0 units and B_y = - 7.0 units. Find the magnitude and direction of the vector C such that A + 2B – 4C = 0.

Answer
A + 2B – 4C = 0
(18 – 6 – 4C_x)i + (- 6 – 14 – 4C_y)j = 0
or C_x = 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.

Problem #7
The vector A has components A_x = -36.0 units and A_y = 24.0 units. Vector B has components B_x = 24.0 units and B_y = 36 units.
[a] Compute the magnitude of A
[b] Compute the magnitude of B
[c] Compute the components of A+ B
[d] Compute the components of 2A - B

Answer
[a] the magnitude of A
The magnitude of A = [(-36)₂+ (24)²]^1/2 = [1872]^1/2 = 43.3 units.
[b] the magnitude of B
 The magnitude of B = [(24)²+ (36)² ]^1/2 = [1872]^1/2 = 43.3 units.
[c] the components of A + B
(A_x + B_x) = - 36.0 + 24.0 = - 12.0 units (A_y + B_y) = 24.0 + 36.0 = 60.0 units
[d] the components of 2A - B
(2A_x - B_x) = -72.0 - 24.0 = - 96.0 units
(2A_y – B_y) = 48.0 – 36.0 = 12.0 units

Problem #8
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.
[b] Express the average velocity of J for the trip from O to d in the unit vector notation.
[c] Determine the magnitude and direction of the average velocity of J for the trip from O to d.
[d] What is the average speed of J for the trip from O to d?

Answer:
[a] Express each displacement vector A, B, C, and D in the usual unit vector notation.
A = - 80.0j m
B = 150 (sin 30⁰i + cos30⁰j) = 75i + 130j m
C = - 50(sin 65⁰i + cos 65⁰j) = -45.3i – 21.1j m
D = 100(- sin 53⁰i + cos 53⁰j) = -80i + 60j m

[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
   = (- 50.3i + 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) = 60.7⁰
[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   

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