# 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?

Feb 16, 2018

His total displacement, $S$ can be written as,the vector sum of two vectors of magnitude $100 m$ 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

Feb 16, 2018

$\text{188 meters}$ (3 significant figures)

#### Explanation:

Using the cosine rule. Let

• $b = 100$
• $c = 100$

The displacement is $a$. The angle between the two $\left(A\right)$ is ${140}^{\circ}$ because the person turns ${40}^{\circ}$ and

${180}^{\circ} - {40}^{\circ} = {140}^{\circ}$

The cosine rule for finding a side is

${a}^{2} = {b}^{2} + {c}^{2} - 2 b c \times C o s A$

Break the formula down into sections:

$2 b c \times C o s A$

$2 \left(100 \times 100\right) = 20000$

$20000 \left(C o s 140\right) = - 15320.88886$

${b}^{2} + {c}^{2}$

${100}^{2} + {100}^{2} = 20000$

So

${b}^{2} + {c}^{2} - 2 b c x C o s A$

$20000 - \left(- 15320.88886\right) = 35320.88886$

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).