Question

Find the resultant of the following displacement. A= 20km 30° south of east, B= 50km due west, C= 40km north east, D= 30km 60° south of west

Answer 1

Answer :

given that,

A = 20 km 30^{0} south of east

B =  50 km due west

C= 40 km north east

D = 30 km 60^{0} south west

We can write in vector form

we have,

$20cos{-30}\widehat{i}+20sin{-30}\widehat{j}$  $-50\widehat{i}$+$40cos{45}\widehat{i} + 40sin{45}\widehat{j}$+$30cos{-60}\widehat{i} + 30sin{-60}\widehat{j}$

Now,

$20\times-\dfrac{\sqrt{3}}{2}\widehat{i} + 20\times\dfrac{1}{2}\widehat{j}-50\widehat{i}+40\times\dfrac{1}{\sqrt{2}}\widehat{i} + 40\times\dfrac{1}{\sqrt{2}}\widehat{j} + 30\times\dfrac{\sqrt{3}}{2}\widehat{i}-30\times\dfrac{1}{2}\widehat{j}$

now, after solving

$(-25\sqrt{3}-50+20\sqrt{2})\widehat{i} + (-25+20\sqrt{2})\widehat{j}$

$(-65.02)\widehate{i} + (3.28)\widehate{j}$

now, the resultant is

$R = \sqrt{(65.02)^{2}+(3.28)^{2}}$

$R= 65.10$

Hence, the resultant is 65.10

Answer 2

Answer:

given that,

A = 20 km 30^{0} south of east

B = 50 km due west

C= 40 km north east

D = 30 km 60^{0} south west

We can write in vector form

we have,

20cos{-30}\widehat{i}+20sin{-30}\widehat{j}20cos−30

i

+20sin−30

j

-50\widehat{i}−50

i

+40cos{45}\widehat{i} + 40sin{45}\widehat{j}40cos45

i

+40sin45

j

+30cos{-60}\widehat{i} + 30sin{-60}\widehat{j}30cos−60

i

+30sin−60

j

Now,

20\times-\dfrac{\sqrt{3}}{2}\widehat{i} + 20\times\dfrac{1}{2}\widehat{j}-50\widehat{i}+40\times\dfrac{1}{\sqrt{2}}\widehat{i} + 40\times\dfrac{1}{\sqrt{2}}\widehat{j} + 30\times\dfrac{\sqrt{3}}{2}\widehat{i}-30\times\dfrac{1}{2}\widehat{j}20×−

2

3

i

+20×

2

1

j

−50

i

+40×

2

1

i

+40×

2

1

j

+30×

2

3

i

−30×

2

1

j

now, after solving

(-25\sqrt{3}-50+20\sqrt{2})\widehat{i} + (-25+20\sqrt{2})\widehat{j}(−25

3

−50+20

2

)

i

+(−25+20

2

)

j

(-65.02)\widehate{i} + (3.28)\widehate{j}(−65.02)\widehatei+(3.28)\widehatej

now, the resultant is

R = \sqrt{(65.02)^{2}+(3.28)^{2}}R=

(65.02)

2

+(3.28)

2

R= 65.10R=65.10