Chapter 20, Problem 1SP

(a)

To determine

The vector diagram to o the velocity of the boat relative to Earth.

Answer to Problem 1SP

The vector diagram is shown is

and the velocity of the boat relative to water is labelled as $v_{\text{sum}}$.

Explanation of Solution

Vector diagram are used to describe the velocity of a moving object during its motion. In a vector diagram, the magnitude of the vector is represented by the length of the vector arrow and the direction is given by the arrow head.

Figure 1

The vector diagram showing the motion of the boat is drawn in figure 1. The velocity of the boat relative to the water is represented by $v_b$ and the velocity of the stream is represented by $v_w$. The velocity of the boat relative to Earth is represented by $v_{\text{sum}}$ according to the triangular law of vector addition.

Conclusion:

Therefore, the vector diagram is shown in the figure 1 and the velocity of the boat relative to water is labelled as $v_{\text{sum}}$.

(b)

To determine

The magnitude of the velocity of boat relative to the Earth.

Answer to Problem 1SP

The magnitude of the velocity of boat relative to the Earth is $6.71\text{m}/\text{s}$.

Explanation of Solution

Given info: The velocity of the boat relative to water is $\text{6 }\text{m}/\text{s}$ and the velocity of the stream is $3\text{m}/\text{s}$.

Write the expression to find the magnitude of the velocity of boat relative to Earth according to Pythagorean Theorem.

$v_{\text{sum}}=\sqrt{v_b^2+v_w^2}$

Here,

$v_{\text{sum}}$ is the velocity of the boat relative to Earth

$v_b$ is the velocity of the boat relative to water

$v_w$ is the velocity of water

Substitute $\text{6 }\text{m}/\text{s}$ for $v_b$ and $3\text{m}/\text{s}$ for $v_w$ in the above equation to find $v_{\text{sum}}$.

$\begin{array}{c}{v}_{\text{sum}}=\sqrt{{\left(\text{6 }\text{m}/\text{s}\right)}^2+{\left(3\text{m}/\text{s}\right)}^2}\\ =6.71\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the magnitude of the velocity of boat relative to the Earth is $6.71\text{m}/\text{s}$.

(c)

To determine

The time taken for the boat to cross the stream.

Answer to Problem 1SP

The time taken for the boat to cross the stream is $8\text{s}$.

Explanation of Solution

Given info: The width of the stream is $48\text{m}$ and the velocity of the boat relative to water is $\text{6}\text{m}/\text{s}$

Write the expression for the time taken for boat to cross the stream.

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

Here,

$d$ is the width of the stream

$v$ is the velocity of the boat

$t$ is the time taken to cross the stream

Substitute $48\text{ m}$ for $d$ and $\text{6 }\text{m}/\text{s}$ for $v$ in the above equation to find $t$.

$\begin{array}{c}t=\frac{48\text{m}}{6\text{m}/\text{s}}\\ =8\text{s}\end{array}$

Conclusion:

Therefore, the time taken for the boat to cross the stream is $8\text{s}$.

(d)

To determine

The distance from starting point till the ending point along the downward flow of the stream.

Answer to Problem 1SP

The distance from starting point till the ending point along the downward flow of the stream is $24\text{ m}$.

Explanation of Solution

Write the expression to find the distance of the between the starting and ending point of the boat along the flow of the stream.

$d=vt$

Here,

$d$ is the distance of the between the starting and ending point

$v$ is the velocity of the water

$t$ is the time taken

Substitute $\text{8 s}$ for $t$ and $3\text{m}/\text{s}$ for $v$ in the above equation to find $d$.

$\begin{array}{c}d=\left(3\text{m}/\text{s}\right)\left(\text{8 s}\right)\\ =24\text{ m}\end{array}$

Conclusion:

Therefore, the distance from starting point till the ending point along the downward flow of the stream is $24\text{ m}$.

(e)

To determine

The distance travelled by the boat in reaching the opposite bank.

Answer to Problem 1SP

The distance travelled by the boat in reaching the opposite bank is $53.7\text{ m}$.

Explanation of Solution

Write the expression to find the magnitude of the resultant of two vectors $d_w$ and $d_b$ using Pythagorean Theorem.

$d_{\text{sum}}=\sqrt{d_b^2+d_w^2}$

Here,

$d_{\text{sum}}$ is the distance travelled by the boat before reaching the opposite bank

$d_b$ is the distance travelled by the boat from starting point to ending point along the river

$d_w$ is the width of the stream

Substitute $24\text{ m}$ for $d_b$ and $48\text{ m}$ for $d_w$ in the above equation to find $d_{\text{sum}}$.

$\begin{array}{c}{d}_{\text{sum}}=\sqrt{{\left(24\text{ m}\right)}^2+{\left(48\text{ m}\right)}^2}\\ =53.7\text{ m}\end{array}$

Conclusion:

Therefore, the distance travelled by the boat before reaching the opposite bank is $53.7\text{ m}$.