Question

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?

Answer 1

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 2

`"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).