Question

deals with the concepts that are important in this problem. A grasshopper makes four jumps. The displacement vectors are (1) 31.0 cm, due west; (2) 32.0 cm, 34.0 ° south of west; (3) 16.0 cm, 57.0 ° south of east; and (4) 16.0 cm, 75.0 ° north of east. Find (a) the magnitude and (b) direction of the resultant displacement. Express the direction as a positive angle with respect to due west.

Answer 1

(a) 46.6 cm

To calculate the total displacement, we need to resolve each vector along the x- and y- direction.

In the following, we take:

- East as positive x-direction, West as negative x-direction

- North as positive y-direction, South as negative y-direction

Vector A:

$A_x = -31.0 cm\\A_y = 0$

Vector B:

$B_x = -(32.0 cm) cos 34.0^{\circ}=-25.6 cm\\B_y = -(32.0 cm) sin 34.0^{\circ} = -17.9 cm$

Vector C:

$C_x = (16.0 cm)cos 57.0^{\circ} =8.7 cm\\C_y = -(16.0 cm)sin 57.0^{\circ} =-13.4 cm$

Vector D:

$D_x = (16.0 cm) cos 75.0^{\circ} =4.1 cm\\D_y = (16.0 cm) sin 75.0^{\circ} =15.5 cm$

So the components of the total displacement are:

$R_x = -31.0 cm - 25.6 cm +8.7 cm +4.1 cm =-43.8 cm\\R_y = 0-17.9 cm -13.4 cm +15.5 cm =-15.8 cm$

So the magnitude of the total displacement is:

$R=\sqrt{R_x^2 + R_y^2}=\sqrt{(-43.8 cm)^2+(-15.8 cm)^2}=46.6 cm$

(b) 19.8 degrees south of west

First of all, let's notice that:

- both components x and y of the total displacement are negative --> this means that the direction of the displacement is south of west

This means that the angle will be given by

$\theta = tan^{-1} (\frac{R_y}{R_x})$

where we have

$R_x = -43.8 cm\\R_y = -15.8 cm$

Substituting, we find

$\theta = tan^{-1} (\frac{-15.8 cm}{43.8 cm})=19.8^{\circ}$

and this angle is south of west.