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?

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Answer 1 · 2018-02-16T14:22:17

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 · 2018-02-16T14:40:54

$"188 meters"$ (3 significant figures)

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).