Chapter 20, Problem 1SP

(a)

To determine

To draw the vector diagram of motion of a boat across stream and the velocity of the boat relative to water.

Answer to Problem 1SP

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

Explanation of Solution

Given info: The velocity of the boat in still water is $7\text{m}/\text{s}$. The velocity of the stream is $3\text{m}/\text{s}$ and the width of the stream is $28\text{ m}$.

Following figure gives the diagram of the vector form of the motion of boat across the figure.

Figure 1

The figure 1 shows the velocity of water as $v_w$ and the velocity of the boat as $v_b$. The resultant of the two vectors gives the velocity of the boat relative to water is labelled as $v_{\text{sum}}$.

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

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 $7.6\text{m}/\text{s}$.

Explanation of Solution

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

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

Here,

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

$v_b$ is the velocity of the boat

$v_w$ is the velocity of water

Substitute $7\text{m}/\text{s}$ for $v_b$ and $3\text{m}/\text{s}$ for $v_w$ in the above equation.

$\begin{array}{c}{v}_{\text{sum}}=\sqrt{{\left(7\text{m}/\text{s}\right)}^2+{\left(3\text{m}/\text{s}\right)}^2}\\ =7.6\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the magnitude of the velocity of boat relative to the Earth is $7.6\text{m}/\text{s}$.

(c)

To determine

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 $4\text{s}$.

Explanation of Solution

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

Substitute $28\text{ m}$ for $d$ and $7\text{m}/\text{s}$ for $v$ in the above equation.

$\begin{array}{c}t=\frac{28\text{ m}}{7\text{m}/\text{s}}\\ =4\text{ s}\end{array}$

Conclusion:

Therefore, the time taken for the boat to cross the stream is $4\text{s}$.

(d)

To determine

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 $12\text{ m}$.

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 $4\text{ s}$ for $t$ and $3\text{m}/\text{s}$ for $v$ in the above equation.

$\begin{array}{c}d=\left(3\text{m}/\text{s}\right)\left(4\text{ s}\right)\\ =12\text{ m}\end{array}$

Conclusion:

Therefore, the distance from starting point till the ending point along the downward flow of the stream is $12\text{ m}$.

(e)

To determine

To determine the distance travelled by the boat before reaching the opposite bank.

Answer to Problem 1SP

The distance travelled by the boat before reaching the opposite bank is $30.5\text{ m}$.

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 to ending point along the river

$d_w$ is the width of the stream

Substitute $12\text{ m}$ for $d_b$ and $28\text{ m}$ for $d_w$ in the above equation.

$\begin{array}{c}{v}_{\text{sum}}=\sqrt{{\left(12\text{ m}\right)}^2+{\left(28\text{ m}\right)}^2}\\ =30.5\text{ m}\end{array}$

Conclusion:

Therefore, the distance travelled by the boat before reaching the opposite bank is $30.5\text{ m}$.