Question

A river flows due south with a speed of 2.00 ${\rm m}/{\rm s}$ . A man steers a motorboat across the river; hisvelocity relative to the water is 4.20 ${\rm m}/{\rm s}$ due east. The river is 800 ${\rm m}$ wide.

a)   What is the magnitude of his velocityrelative to the earth?

         v=     m/s

b)   What is the direction of his velocityrelative to the earth?

         θ=       ^o south ofeast

c)   How much time is required for the man tocross the river?

        t=      s

d)   How far south of his starting point willhe reach the opposite bank?

         d=     m

Answer 1

Concepts and reason

The main concept used is the magnitude and the direction of the vector.

Initially, draw the diagram for the system and form the diagram specify $x{\rm{ and }}y$ component of the system. Finally, this component is responsible for the calculation of the magnitude and the direction of the system.

Fundamentals

The expression of the magnitude for $x{\rm{ and }}y$ component is equal to,

$r = \sqrt {{x^2} + {y^2}}$
‎Here, $r$ is the magnitude.

The expression of the direction of the $x{\rm{ and }}y$ is equal to,

$\theta = {\tan ^{ - 1}}\left( {\frac{y}{x}} \right)$

Here, $\theta$ is the angle between the component $x{\rm{ and }}y$ .

(a)

The diagram for the provided system is given below,

From the above diagram, it is clear the velocity in $x$ and $y$ component is equal to,

Velocity in $y$ direction is equal to, ${v_y} = 2.0{\rm{ m/s}}$

Velocity in $x$ direction is equal to, ${v_x} = 4.2{\rm{ m/s}}$

The expression of the magnitude of his velocity relative to the earth is equal to,

$v = \sqrt {v_x^2 + v_y^2}$

Here, ${v_x}$ is the velocity in $x$ component, ${v_y}$ is the velocity in $y$ component and $v$ is the resultant velocity.

Substitute $2.0{\rm{ m/}}{{\rm{s}}^2}$ for ${v_y}$ and $4.2{\rm{ m/}}{{\rm{s}}^2}$ for ${v_x}$ in the above expression of the resultant velocity,

$\begin{array}{c}\\v = \sqrt {{{\left( {4.2{\rm{ m/}}{{\rm{s}}^2}} \right)}^2} + {{\left( {2.0{\rm{ m/}}{{\rm{s}}^2}} \right)}^2}} \\\\ = 4.65{\rm{ m/s}}\\\end{array}$

(b)

Calculation of direction of his velocity relative to the earth.

The expression of the direction in terms of the velocity is equal to,

$\theta = {\tan ^{ - 1}}\left( {\frac{{{v_y}}}{{{v_x}}}} \right)$

Substitute $2.0{\rm{ m/}}{{\rm{s}}^2}$ for ${v_y}$ and $4.2{\rm{ m/}}{{\rm{s}}^2}$ for ${v_x}$ in the above expression of the resultant velocity,

$\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{2.0{\rm{ m/}}{{\rm{s}}^2}}}{{4.2{\rm{ m/}}{{\rm{s}}^2}}}} \right)\\\\ = 25.46^\circ \\\end{array}$

(c)

Calculate the time required for the man to cross the river.

The expression of the distance in terms of the speed is equal to,

$s = vt$

Rearrange the above expression in terms of the time $t$ .

$t = \frac{s}{v}$

Substitute $800{\rm{ m}}$ for $s$ and $4.2\;{\rm{m/s}}$ for $v$ in the above expression of the time,

$\begin{array}{c}\\t = \frac{{800{\rm{ m}}}}{{4.2\;{\rm{m/s}}}}\\\\ = 190.47{\rm{ s}}\\\end{array}$

(d)

Calculate the starting point while reach the opposite bank of the river.

The expression of the distance while reach the opposite bank of the river is equal to,

$s = vt$

Substitute $2.0{\rm{ m/s}}$ for $v$ and $190.47{\rm{ s}}$ for $t$ in the above expression of the distance $s$

$\begin{array}{c}\\s = \left( {2.0{\rm{ m/s}}} \right)\left( {190.47{\rm{ s}}} \right)\\\\ = 380.94{\rm{ m}}\\\end{array}$

Ans: Part a

The magnitude of his velocity relative to the earth is equal to $4.65{\rm{ m/s}}$ .

Part b

The direction of his velocity relative to the earth is equal to $25.46^\circ$ .

Part c

The time required for the man to cross the river is equal to $190.47{\rm{ s}}$ .

Part d

The starting point while reach the opposite bank of the river is equal to $380.94{\rm{ m}}$ .