A person walked 100 meters towards north and then took a turn of 40° to his left and walked another 100 meters,what is his total displacement?

2 Answers
Feb 16, 2018

His total displacement, #S# can be written as,the vector sum of two vectors of magnitude #100m # making an angle of #40# in between them.

So,we can calculate the displacement as,

# S = sqrt( 100^2 + 100^2 + 2×100×100×cos 40) = 187.94 m#

Answer:

#"188 meters"# (3 significant figures)

Explanation:

Using the cosine rule. Let

  • #b=100#
  • # c=100#

The displacement is #a#. The angle between the two #(A)# is #140^@# because the person turns #40^@# and

#180^@ - 40^@ = 140^@#

The cosine rule for finding a side is

#a^2 = b^2 + c^2 - 2bc xx CosA#

Break the formula down into sections:

#2bc xx CosA#

#2(100 xx 100) = 20000#

#20000(Cos140) = -15320.88886#

#b^2 + c^2#

#100^2 + 100^2 = 20000#

So

#b^2 + c^2 - 2bc x CosA#

#20000 - (-15320.88886) = 35320.88886#

Therefore, we know that

#a^2 = 35320.88886#

We now must find the square root to find what #a# (the displacement) is. Using a calculator,

#sqrt(35320.88886) = 187.938524#

We can round this to #188# meters (#3# significant figures).