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

(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°

Step-by-step 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=√(y^2+x^2)`

Put the value into the formula

`D=√((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)((y)/(x))`

Put the value into the formula

`\theta=\tan^(-1)((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°