Question

A person swims 6.4 meters per second north while being pushed by a current moving west at 2.1 meters per second. What is the direction of the swimmer's resultant vector? Hint: Draw a vector diagram. Ө 0 = [ ? ]° Round your answer to the nearest hundredth.

Answer 1

The swimmer's resultant vector can be found by drawing a right triangle with the northward velocity as one leg and the westward velocity as the other leg. The hypotenuse of this triangle represents the swimmer's resultant velocity. The angle Ө0 between the northward velocity and the resultant velocity can be found using the inverse tangent function: tan⁻¹(2.1/6.4) = 18.19°. So, the direction of the swimmer's resultant vector is 18.19° west of north.

Answer 2

108.17° (nearest hundredth)

Explanation:

In order to find the direction the person is swimming, we must find the direction of the resultant vector of the two vectors representing 6.4 m/s north and 2.1 m/s west, measured counterclockwise from the positive x-axis.

Since the two vectors form a right angle, we can use the tangent trigonometric ratio.

`\boxed{\begin{minipage}{7 cm}\underline{Tangent trigonometric ratio} \\\\$ \tan x=(O)/(A)$\\\\where:\\ \phantom{ww}$\bullet$ $x$ is the angle. \\ \phantom{ww}$\bullet$ $\sf O$ is the side opposite the angle. \\\phantom{ww}$\bullet$ $\sf A$ is the side adjacent the angle.\\\end{minipage}}`

The resultant vector is in quadrant II, since the swimmer is travelling north (positive y-direction) and is being pushed by a current moving west (negative x-direction).

As the direction of a resultant vector is measured in an anticlockwise direction from the positive x-axis (and the resultant vector is in quadrant II), we need to add 90° to the angle found using the tan ratio.

The angle between the y-axis and the resultant vector can be found using tan x = 2.1 / 6.4. Therefore, the expression for the direction of the resultant vector θ is:

`\theta=90^(\circ)+\arctan \left((2.1)/(6.4)\right)`

`\theta=90^(\circ)+18.1659565...^(\circ)`

`\theta=108.17^(\circ)\; \sf (nearest\;hundredth)`

Therefore, the direction of the swimmer's resultant vector is approximately 108.17° (measured anticlockwise from the positive x-axis).

This can also be expressed as N 18.17° W.