Question

A drunken sailor stumbles 600 meters north, 550 meters northeast, then 500 meters northwest. What is the total displacement and the angle of the displacement? (Assume east is the +x-direction and north is the +y-direction.)

a. magnitude
b. direction
c. ° counterclockwise from the +x-axis

Answer 1

Answer:

(a). the magnitude of displacement is 1342.92 m.

(b). The direction is east of north 88.4°.

(c). The angle of counterclockwise from x-axis is 88.4°

Explanation:

Given that,

A drunken sailor stumbles 600 meters north, 550 meters northeast, then 500 meters northwest.

In x- direction,

$x=550\cos45-500\cos45$

$x=35.35\ m$

In y-direction,

$y=600+550\sin45+500\sin45$

$y=1342.46\ m$

(a). We need to calculate the magnitude of displacement

Using formula of displacement

$D=\sqrt{y^2+x^2}$

Put the value into the formula

$D=\sqrt{(600+550\sin45+500\sin45)^2+(550\cos45-500\cos45)^2}$

$D=1342.92\ m$

(b). We need to calculate the direction

Using formula of direction

$\theta=\tan^{-1}(\dfrac{y}{x})$

Put the value into the formula

$\theta=\tan^{-1}(\dfrac{1342.46}{35.35})$

$\theta=88.4^{\circ}$

The direction is east of north 88.4°.

(c). The angle of counterclockwise from x-axis is 88.4°

Hence, (a). the magnitude of displacement is 1342.92 m.

(b). The direction is east of north 88.4°.

(c). The angle of counterclockwise from x-axis is 88.4°