Chapter 6.3, Problem 36SSC

To determine

The velocity of M relative to the water.

Answer to Problem 36SSC

  $3.85\text{m/s}$ , $9.26°$ north of east.

Explanation of Solution

Given:

Boat is travelling east at $3.8\text{m/s}$ while M moves north across the deck of deck at $0.62\text{ m/s}$ .

Formula used:

Relative velocity of an object a to object c is given by

  $v_{a/c}=v_{a/b}+v_{b/c}$

Calculation:

Assume the east direction to be horizontal direction towards right, then north direction will be vertically upwards direction.

Now, the velocity of M relative to water can be written as,

  $v{\text{Martin/water}}_{}=v_{\text{Martin/boat}}+v_{\text{boat/water}}$

Since the above velocities are at right angle, thus M’s velocity relative to water can be written as,

  $v{\text{Martin/water}}_{}=\sqrt{{\left(v_{\text{Martin/boat}}\right)}^2+{\left(v_{\text{boat/water}}\right)}^2}$

Using given values in above,

  $\begin{array}{c}v{\text{Martin/water}}_{}=\sqrt{{\left(0.62\right)}^2+{\left(3.8\right)}^2}\\ v{\text{Martin/water}}_{}=\sqrt{0.3844+14.44}\\ v{\text{Martin/water}}_{}=3.85\text{m/s}\end{array}$

At an angle,

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{v_{\text{Martin/boat}}}{v_{\text{boat/water}}}\right)\\ ={\tan }^{-1}\left(\frac{0.62}{3.8}\right)\\ =9.26°\text{ with respect to the East}\end{array}$

Conclusion:

Thus, velocity of the M is $3.85\text{m/s}$ , $9.26°$ north of east.