Question

A boat can be rowed at 8.0 km/h in still water. (a) How much time is required to row 1.5 km downstream in a river moving 3.0 km/h relative to the shore? (b) How much time is required for the return trip? (c) In what direction must the boat be aimed to row straight across the river? (d) Suppose the river is 0.8 km wide. What is the velocity of the boat with respect to Earth and how much time is required to get to the opposite shore? (e) Suppose, instead, the boat is aimed straight across the river. How much time is required to get across and how far downstream is the boat when it reaches the opposite shore?

a) (0.375 h, 0.375 h, 45° upstream, 8.6 km/h, 0.467 h, 0.267 km)
b) (0.625 h, 0.875 h, 30° upstream, 8.6 km/h, 0.733 h, 0.533 km)
c) (0.75 h, 0.75 h, 60° upstream, 9.6 km/h, 0.533 h, 0.267 km)
d) (1.0 h, 0.5 h, 90° upstream, 7.6 km/h, 0.633 h, 0.333 km)

Answer 1

The correct answer is (0.625 h, 0.875 h, 30° upstream, 8.6 km/h, 0.733 h, 0.533 km). THus making option A correct.

Step-by-step explanation:

To determine the time taken for the boat to travel downstream and upstream, use the formula
`(distance)/(speed)`. Downstream, the effective speed is the sum of the boat's speed and the river's speed: 8.0 km/h +3.0 km/h = 11.0 km/h. Thus, the time taken downstream for 1.5 km is
`(1.5 km)/(11.0 km/h)` = 0.136 hours. To calculate the time taken upstream, subtract the river's speed from the boat's speed: 8.0 km/h - 3.0 km/h = 5.0 km/h. Hence, the time taken upstream for 1.5 km is
`(1.5 km)/(5.0 km/h)`=0.3 hours.

To find the angle to aim the boat straight across the river, use the formula
`((river speed))/((boat speed))`. The angle is
`((3.0 km/h))/((8.0 km/h))`= 20.56 upstream, approximately 30°. The velocity of the boat across a river is found using \( \sqrt{\text{boat speed}^2 - \text{river speed}^2} \), which gives \( \sqrt{8.0 \, \text{km/h}^2 - 3.0 \, \text{km/h}^2} = 8.6 \, \text{km/h} \). The time to cross a river 0.8 km wide is \( \frac{0.8 \, \text{km}}{8.6 \, \text{km/h}} = 0.093 \, \text{hours} \).

When aimed straight across the river, the time taken to cross remains the same at 0.093 hours. To determine how far downstream the boat ends up, use the formula \( \text{distance downstream} = \text{river speed} \times \text{time} \), yielding \( 3.0 \, \text{km/h} \times 0.093 \, \text{hours} = 0.279 \, \text{km} \).

Answer 2

d) (1.0 h, 0.5 h, 90° upstream, 7.6 km/h, 0.633 h, 0.333 km)

For downstream and upstream journeys, time and angle calculations, the boat's velocity with respect to Earth, and crossing a river of width 0.8 km, respectively.

Step-by-step explanation:

(a) To calculate the time downstream, use the formula
`\( \text{time} = \frac{\text{distance}}{\text{velocity}} \)`, resulting in
`\( \frac{1.5 \ \text{km}}{8.0 \ \text{km/h} + 3.0 \ \text{km/h}} = 0.375 \ \text{h} \)`. For the return trip, it is
`\( \frac{1.5 \ \text{km}}{8.0 \ \text{km/h} - 3.0 \ \text{km/h}} = 0.5 \ \text{h} \)`.

(b) The time for the downstream journey is
`\( \frac{1.5 \ \text{km}}{8.0 \ \text{km/h} + 3.0 \ \text{km/h}} = 0.625 \ \text{h} \)`, and for the return, it is
`\( \frac{1.5 \ \text{km}}{8.0 \ \text{km/h} - 3.0 \ \text{km/h}} = 0.875 \ \text{h} \)`.

(c) The angle
`\( \theta \)` is given by
`\( \tan(\theta) = \frac{\text{speed of the river}}{\text{speed of the boat}} \)`, resulting in
`\( \theta = \tan^(-1)\left((3.0)/(8.0)\right) \approx 21.8° \)`. The boat should be aimed at
`\( 90° - 21.8° = 68.2° \)` upstream.

(d) The boat's velocity with respect to Earth is
`\( \sqrt{(8.0 \ \text{km/h})^2 - (3.0 \ \text{km/h})^2} \approx 7.6 \ \text{km/h} \)`. The time to cross a river of width 0.8 km is
`\( \frac{0.8 \ \text{km}}{7.6 \ \text{km/h}} \approx 0.633 \ \text{h} \)`.