Question

A boat can travel$2.20\mathrm{m}/\mathrm{s}$in still water. (a) If the boat points directly across a stream whose current is$1.20\mathrm{m}\mathbf{/}\mathbf{s}$, what is the velocity (magnitude and direction) of the boat relative to the shore? (b) What will be the position of the boat, relative to its point of origin, after 3.00 s?

Short answer

(a) The velocity of the boat with respect to the shore is $2.51\mathrm{m}/\mathrm{s}$, and the angle of the velocity of the boat relative to the shore is $61.39°$(as measured in the anticlockwise direction from the horizontal axis).

(b) The boat will be at a distance of 7.53 m, and at an angle of $61.39°$(as measured in the anticlockwise direction from the horizontal axis)

Step-by-step solution

Step 1. Understanding the concept of relative velocity

Relative velocity is equal to the vector difference between the velocity vectors of both objects.

Step 2. Given data and assumptions

Let us assume that the direction of the water current is toward the positive x-axis, and the motion of the boat is along the y-axis.

The velocity vector of the water current with respect to the shore,${\overrightarrow{v}}_{\mathrm{w}}=\left(1.20\mathrm{m}/\mathrm{s}\right)\hat{i}$

The velocity vector of the boat with respect to the water current,${\overrightarrow{v}}_{\mathrm{b}/\mathrm{w}}=\left(2.20\mathrm{m}/\mathrm{s}\right)\hat{j}$

Let the velocity vector of the boat with respect to the shore be represented as ${\overrightarrow{v}}_{\mathrm{b}/\mathrm{s}}$.

Step 3. Vector diagram for different velocity vectors

The vector diagram representing all the velocity vectors can be drawn as follows:

Step 4. Calculating the velocity of the boat with respect to the shore

From the above vector diagram, the value of vector ${\overrightarrow{v}}_{\mathrm{b}/\mathrm{s}}$can be written as:

${\overrightarrow{v}}_{\mathrm{b}/\mathrm{s}}={\overrightarrow{v}}_{\mathrm{w}}+{\overrightarrow{v}}_{\mathrm{b}/\mathrm{w}}$

Substituting the values in the above equation, you get:

${\overrightarrow{v}}_{\mathrm{b}/\mathrm{s}}=\left(1.20\mathrm{m}/\mathrm{s}\right)\hat{i}+\left(2.20\mathrm{m}/\mathrm{s}\right)\hat{j}$

The magnitude of vector ${\overrightarrow{v}}_{\mathrm{b}/\mathrm{s}}$will be equal to the square root of the sum of its components along the axes. Thus,

$\begin{array}{c}\left|{\overrightarrow{v}}_{\mathrm{b}/\mathrm{s}}\right|=\sqrt{{1.20}^2+{2.20}^2}\\ =2.51\mathrm{m}/\mathrm{s}\end{array}$

The tangent function of ${\overrightarrow{v}}_{\mathrm{b}/\mathrm{s}}$with respect to the positive direction of the x-axis can be written as:

$\begin{array}{c}\tan \theta =\frac{\left|{\overrightarrow{v}}_{\mathrm{b}/\mathrm{w}}\right|}{\left|{\overrightarrow{v}}_w\right|}\\ =\frac{2.20}{1.20}\end{array}$

The above equation can be written as:

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{2.20}{1.20}\right)\\ =61.39°\end{array}$

Thus, the velocity of the boat with respect to the shore is $2.51\mathrm{m}/\mathrm{s}$, and it is along the direction $61.39°$, measured an anticlockwise manner from the horizontal axis.

Step 5. Calculating the location of the boat after 3.00 s

The distance traveled (d) by the boat in these 3.00 seconds can be written as:

$\begin{array}{c}d=\left|{\overrightarrow{v}}_{\mathrm{b}/\mathrm{s}}\right|\times 3\\ =2.51\times 3\\ =7.53\mathrm{m}\end{array}$

The direction of displacement of the boat from its initial position will be the same as the direction of ${\overrightarrow{v}}_{\mathrm{b}/\mathrm{s}}$.

Thus, the boat will be at a distance of 7.53 m, and at an angle of $61.39°$with the positive x-axis in an anticlockwise manner.