G Vector Sheet: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | Go up
 - by Chris Murray, But Mostly Brad Hamlin 2002

1. Convert this vector to Vector Component form:                        

The first thing to do is draw the components:

The blue component is the x component, and the red is the y component.

The x component (blue) in this case is to the right, which is the positive direction for x, and is the side adjacent to the known angle, so we use cosine to find it:

v_x = (15 m/s)cos(29^o) = 13.12 m/s
 
The y component (red) in this case is up, which is the positive direction for y, and is the side opposite to the known angle, so we use sine to find it:

v_y = (15 m/s)sin(29^o) = 7.27 m/s
 
And finally we can write it in VC notation:

v = 13.12 m/s x + 7.27 m/s y
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2. Convert this vector to Vector Component form:   

The first thing to do is draw the components:

The blue component is the x component, and the red is the y component.

The x component (blue) in this case is to the right, which is the positive direction for x, and is the side adjacent to the known angle, so we use cosine to find it:

s_x = (300 m)cos(40^o) = 230 m
 
The y component (red) in this case is up, which is the positive direction for y, and is the side opposite to the known angle, so we use sine to find it:

s_y = (300 m)sin(40^o) = 193 m
 
And finally we can write it in VC notation:

s = 230 m x + 193 m y
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3. Convert this vector into vector component notation:

The first thing to do is draw the components:

The blue component is the x component, and the red is the y component.

The x component (blue) in this case is to the right, which is the positive direction for x, and is the side opposite to the known angle, so we use sine to find it:

v_x = (32.5 m/s)sin(55^o) = 18.6 m/s
 
The y component (red) in this case is down, which is the negative direction for y, and is the side adjacent to the known angle, so we use cosine to find it:

v_y = (32.5 m/s)cos(55^o) = 26.6 m/s
 
And finally we can write it in VC notation:

v = 18.6 m/s x - 26.6 m/s y

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4. Convert this vector into vector component notation:

The first thing to do is draw the components:

The blue component is the x component, and the red is the y component.

The x component (blue) in this case is to the left, which is the negative direction for x, and is the side opposite to the known angle, so we use sine to find it:

v_x = (120 m/s)sin(38^o) = 73.8 m/s
 
The y component (red) in this case is down, which is the negative direction for y, and is the side adjacent to the known angle, so we use cosine to find it:

v_y = (15 m/s)cos(38^o) = 94.6 m/s
 
And finally we can write it in VC notation:

v = -73.8 m/s x - 94.6 m/s y

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5. Convert this vector into vector component notation: 

The first thing to do is draw the components:

The blue component is the x component, and the red is the y component.

The x component (blue) in this case is to the left, which is the negative direction for x, and is the side adjacent to the known angle, so we use cosine to find it:

s_x = (132 m)cos(18^o) = 126 m
 
The y component (red) in this case is up, which is the positive direction for y, and is the side opposite to the known angle, so we use sine to find it:

s_y = (132 m)sin(18^o) = 40.8 m
 
And finally we can write it in VC notation:

s = -126 m x + 40.8 m y

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6. Convert this vector into vector component notation: 

The first thing to do is draw the components:

The blue component is the x component, and the red is the y component.

The x component (blue) in this case is to the left, which is the negative direction for x, and is the side adjacent to the known angle, so we use cosine to find it:

a_x = (160 m/s/s)cos(30^o) = -140 m/s/s
 
The y component (red) in this case is down, which is the negative direction for y, and is the side opposite to the known angle, so we use sine to find it:

a_y = (160 m/s/s)sin(30^o) = 80 m/s
 
And finally we can write it in VC notation:

v = -140 m/s/s x - 80. m/s/s y

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7. Convert 5.0 m x + 6.0 m y to Angle Magnitude notation:

The first thing to do is to draw the vector itself:

Here, the blue is the x component, and the red is the y.  The black vector is the vector sum of the two or "the vector"

The magnitude of the vector is the hypotenuse of the triangle:

hyp = Ö{(5.0 m)² + (6.0 m)²} = 7.81 m = 7.8 m

The angle between the x axis (blue) and the vector (black) is the inverse tangent of opposite over adjacent:

q =Tan^-1{6.0 m/5.0 m} = 50.2^o
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8. Convert -3.00m x+ 7.00m y into angle magnitude notation: 

First, we draw the vector.

Here, the blue is the x component, and the red is the y component.  The black is the actual vector that we are trying to find.

The magnitude of that vector is the hypotenuse of the triangle, so Ö{3²+9²}= 7.62m.

The angle between the x- axis and the vector is simply the inverse tangent of opposite over adjacent.

q =Tan^-1{7.00 m/3.00 m} = 66.8^o 
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9. Convert -4.2m x - 3.2m y into angle magnitude notation 

First, we draw the vector 


Here, blue is the x- component, and red is the y- component. The black is our actual vector.

The magnitude of the vector is the hypotenuse of the triangle, so Ö{3.2²+4.2²}= 5.3m

The angle between the x- axis and the vector is simply the inverse tangent of opposite over adjacent.

q =Tan^-1{3.2 m/4.2 m} = 37^o 
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10. Convert 1.12m x - 5.70m y into angle magnitude notation 

First, we draw the vector


Here, blue is the x- component, and red is the y- component. The black is our actual vector.

The magnitude of the vector is the hypotenuse of the triangle, so Ö{1.12²+5.70²}= 5.81m

The angle between the x- axis and the vector is simply the inverse tangent of opposite over adjacent.

q =Tan^-1{5.70 m/1.12 m} = 78.9^o 

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11. Add vectors #7 & #8 from part #2. 

5.0m x	+	6.0m y
+ -3.00m x	+	7.00m y
2.0m x	+	13.0m y

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12. Subtract Vector #8 from #7.

5.0m x	+	6.0m y
+-(-3.00m) x	+	-(7.00m) y
8.0m x	-	1.0m y

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13. Subtract vector #8 from vector #9.

-4.2m x	-	3.2m y
+-(-3.00m) x	+	-(7.00m) y
-1.2m x	-	10.2m y

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14. Add vectors #10 & #9.

1.12m x	-	5.70m y
-4.2m x	-	3.2m y
-3.1m x	-	8.9m y

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15. Add these angle magnitude vectors analytically, and express their sum as an angle magnitude vector.

A:                      B:

First, we convert them into vector component vectors, then add them, and then convert them back into angle magnitude vectors.
So, step one:

A:

The blue component is the x component, and the red is the y component.

The x component (blue) in this case is to the right, which is the positive direction for x, and is the side adjacent to the known angle, so we use cosine to find it:

s_x = (12 m)cos(25^o) = 10.9 m
 
The y component (red) in this case is up, which is the positive direction for y, and is the side opposite to the known angle, so we use sine to find it:

s_y = (12 m)sin(25^o) = 5.1 m
 
And finally we can write it in VC notation:

A = 10.9 m x + 5.1 m y

B:

The blue component is the x component, and the red is the y component.

The x component (blue) in this case is to the right, which is the positive direction for x, and is the side adjacent to the known angle, so we use cosine to find it:

s_x = (10. m)cos(48^o) = 6.69 m
 
The y component (red) in this case is down, which is the negative direction for y, and is the side opposite to the known angle, so we use sine to find it:

s_y = (10.m)sin(48^o) = 7.43 m
 
And finally we can write it in VC notation:

B = 6.7 m x - 7.4 m y

So now we can add A & B:

10.9m x	+	5.1m y
+6.7m x	-	7.4m y
17.6m x	-	2.3m y

Then, we change that vector component vector back into an angle magnitude vector.



Here, the blue is the x component, and the red is the y.  The black vector is the vector sum of the two or "the vector"

The magnitude of the vector is the hypotenuse of the triangle:

hyp = Ö{(17.6 m)² + (-2.3 m)²} =  17.74m = 18 m

The angle between the x axis (blue) and the vector (black) is the inverse tangent of opposite over adjacent:

q =Tan^-1{2.3 m/17.6 m} = 7.7<sup>o

So, the answer is 18m, 7.7 degrees below the x-axis.</sup>
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16. Add these two vectors analytically, and express their sum as an Angle-Magnitude vector.

A:                  B:

First, we convert them into vector component vectors, then add them, and then convert them back into angle magnitude vectors.

A:

The blue component is the x component, and the red is the y component.

The x component (blue) in this case is to the right, which is the positive direction for x, and is the side adjacent to the known angle, so we use cosine to find it:

s_x = (32 m)cos(30^o) = 27.7 m
 
The y component (red) in this case is up, which is the positive direction for y, and is the side opposite to the known angle, so we use sine to find it:

s_y = (32 m)sin(30^o) = 16 m
 
And finally we can write it in VC notation:

A = 27.7 m x + 16 m y

B:

The blue component is the x component, and the red is the y component.

The x component (blue) in this case is to the right, which is the positive direction for x, and is the side opposite to the known angle, so we use sine to find it:

s_x = (40. m/s)sin(15^o) = 10.35 m
 
The y component (red) in this case is down, which is the negative direction for y, and is the side adjacent to the known angle, so we use cosine to find it:

s_y = (40. m/s)cos(15^o) = 38.64 m
 
And finally we can write it in VC notation:

B = 10.4 m x - 38.6 m y

So now we can add A & B:

27.7m x	+	16m y
+10.4m x	-	38.6m y
38.1m x	-	22.6m y

Then, we change that vector component vector back into an angle magnitude vector.

Here, the blue is the x component, and the red is the y.  The black vector is the vector sum of the two or "the vector"

The magnitude of the vector is the hypotenuse of the triangle:

hyp = Ö{(38.1 m)² + (22.6 m)²} =  44.29m = 44 m

The angle between the x axis (blue) and the vector (black) is the inverse tangent of opposite over adjacent:

q =Tan^-1{22.6m/38.1 m} = 30.67<sup>o

So, the answer is 44m, 31 degrees below the x-axis.</sup>
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