VECTORS TEST II SOLUTIONS

Adding or subtraction of vectors. Multiplying or dividing of vectors by a scalar number. Separated of vectors into its components. Solutions of the questions.



Solution 1

The vector K is 3 units. If we multiply by 2, it is be 6 units.

The vector N is one unit. If we multiply by – 4, its magnitude would be 4 unit and its direction is inverted.

Magnitude of the vector M is 3 units and its direction is West. If we multiply by (– 3) its magnitude would be 9 units, and its direction is would be East.

No changes are do on the vector L.

Because of length of each square is 30 N, magnitudes of these vectors are as follows.

2K = 6•30 = 180 N

4N = 4√2•30 = 120√2 N

L = 3•30 = 90 N

3M = 9•30 = 270 N

If we add the vectors as in figure, we get the resultant R vector.

Vectors-t2c1


 

Magnitude of the vector R can found using the Pythagorean Theorem. The vector R is the hypotenuse of a right triangle which its perpendicular edges are 5 units.

The magnitude of the vector R is 5√2 unit squares. Because of the length of each square is 30 N,


R = 5√2•30 = 150√2 N


The right answer is E option.


Solution 2

We do not any changes on the K, M and P vectors.

We reverse the vector L and multiply by 2.

We reverse the vector N and multiply by 3.

If we add these vectors as in the figure, we obtain the resultant vector R.

The magnitude of the vector R can found using the Pythagorean Theorem.

The vector R is hypotenuse of a right triangle whose edges are 3 and 4 units.

 Vectors-t2c2


R2 = 32 + 42

R2 = 9 + 16

R = 5

Because of the length of each unit square is 25 N,

R = 5•25 = 125 N.


The right answer is C option.


Solution 3

 

We multiply the vector L by 5. We multiply the vector K by 3 and inverted it.

We divide the vector M by 2.

We don’t any change on the vector P and N.

If we add these vectors by head-to-tail method, we get the resultant vector R.

Vectors-t2c3


The vector R is in the West direction and has a size of 5 units.


The right answer is B option.


Solution 4 

 Vectors-t2c4


If we add the vectors of flow and the boat by head-to-tail method, we create a right triangle. Hypotenuse of this triangle is the velocity vector of the boat in the river.

We can find hypotenuse of the right triangle using the Pythagorean Theorem.

R2 = 32 + 42

R = 5 m/s


Magnitude of the velocity vector of the boat in the river is 5 m/s. This velocity vector is sum of the flow rate vector and the boat velocity vector.


The right answer is A option.


Solution 5 

 Vectors-t2c5


Firstly, let we find, time of arrival of the boat to the opposite coast. This duration is time of trip of the boat in the river.


Time of arrival of the boat to the opposite coast.

t = 360/3 = 120 s.


The boat is has been stayed in the river for 2 seconds. The boat is has been moved downstream by the river duration 2 seconds.

The distance the boat takes to downward,

X = 120 • Va

X = 120•2

X = 240 m


The right answer is C option.


Solution – 6

Vectors-t2c6


We will separate the vectors into its components and we will add the components.

Ax = 200•cos30°

Ax = 200•0.87

Ax = 174 N


Ay = 200•sin30°

Ay = 200•0.5

Ay = 100 N


Bx = -100.cos60°

Bx = -50 N


By = -100.sin60

By = -87 N


Now, let we add on the same axis.

Rx = Ax + Bx

Rx = 174 – 50 

Rx = 124 N


Ry = Ay + By

Ry = 100 – 87

Ry = 13 N

Now, we have two vectors that perpendicular to each other. The resultant vector R can found using the Pythagorean Theorem.

Vectors-t2c6b


R = 1242 + 132

R ≅ 41 N


The right answer is D option.


Vector Questions 2

Vector Questions 1


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26/09/2018

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