Question

A car is driven 250 km west and then 98 kmsouthwest (45∘).What is the displacement of the car from the point of origin (magnitude)?What is the displacement of the car from the point of origin (direction)?

Answer 1

First position in component form

$A_x = -250 km \widehat{i}$

$A_y = 0 km \widehat{j}$

Second position in component form

$B_x=- 98 km \cos 45 \widehat{i} \Rightarrow - 67.18 km \widehat{i}$

$B_y= -98 km \sin 45 \widehat{j} \Rightarrow - 67.18 km \widehat{j}$

Displacement is

$C_x = A_x + B_x=- 250 \widehat{i} - 67.18 km \Rightarrow -317.18 \widehat{i}$

$C_y = A_y + B_y= 0-67.18 km \widehat{j} \Rightarrow - 67.18 km \widehat{j}$

Magnitude is

$C = \sqrt{C_x^2 + C_y^2} = \sqrt{(-317.18)^2 + (-67.18)^2} = 324.21 km$

Direction is

$\theta = \tan^{-1} \left ( \frac{-67.18}{-317.18} \right ) \Rightarrow 11.95^o$

The direction is 11.95 below the negative x axis or 191.95 from positive x axis.