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Chapter 3: Problem 35

A river flows due south with a speed of 2.0 m/s. You steer a motorboat across the river; your velocity relative to the water is 4.2 m/s due east. The river is 500 m wide. (a) What is your velocity (magnitude and direction) relative to the earth? (b) How much time is required to cross the river? (c) How far south of your starting point will you reach the opposite bank?

Short Answer

Expert verified
(a) Velocity: 4.66 m/s, 25.0掳 south of east; (b) Time: 119.05 seconds; (c) Distance south: 238.1 meters.

Step by step solution

01

Determine the Boat's Velocity Relative to Earth

The velocity of the boat relative to water is given as 4.2 m/s due east, and the velocity of the river is 2.0 m/s due south. To find the resultant velocity of the boat relative to the earth, we need to combine these two perpendicular velocities using the Pythagorean theorem. The magnitude of the resultant velocity, \( v \), is calculated as: \[ v = \sqrt{(4.2)^2 + (2.0)^2} \approx 4.66 \text{ m/s} \] To find the direction, use the arctangent function: \( \theta = \tan^{-1}\left(\frac{2.0}{4.2}\right) \approx 25.0^\circ \) south of east.
02

Calculate the Time to Cross the River

The time \( t \) to cross the river can be calculated using the width of the river and the component of velocity perpendicular to the river's flow. Since the boat's speed relative to water is 4.2 m/s directly east (perpendicular to the river), the time is: \[ t = \frac{500}{4.2} \approx 119.05 \text{ seconds} \]
03

Calculate How Far South the Boat will Drift

The boat's velocity downstream (due south) is 2.0 m/s. The distance drifted south \( d \) is given by the formula: \( d = \text{velocity} \times \text{time} \). Using the time calculated in the previous step: \[ d = 2.0 \times 119.05 \approx 238.1 \text{ meters} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

River Crossing
A river crossing is more than just steering from one side to the other. It involves understanding how different forces and velocities interact. When crossing a river with a flowing current, the path you take will be affected by the velocity of the water. If you steer a boat directly across, the current will push you downstream. The key is to consider both your intended path and the river's flow. This ensures that you know where you'll end up, relative to where you started. Think of it as combining two paths: the one you steer and the natural path of the river.
Resultant Velocity
Resultant velocity is the combined effect of two or more individual velocities. In river crossing problems, it refers to the velocity of the boat relative to the earth. Here, we consider the velocity of the boat and the velocity of the river. To find the resultant, these velocities need to be combined. This requires special attention when they are perpendicular, like in this case where the boat moves east (4.2 m/s) and the river moves south (2.0 m/s). Combine them using vector addition, often solved using the Pythagorean theorem.
Pythagorean Theorem
The Pythagorean theorem is a mathematical tool that helps in calculating the magnitude of the resultant velocity when two velocities are at right angles. For a right triangle where the two legs represent the known velocities (east and south), the hypotenuse gives the total velocity. The formula is:
  • \[ v = \sqrt{(\text{velocity}_{east})^2 + (\text{velocity}_{south})^2} \]
This provides not just the speed, but a complete picture when combined with direction, calculated using arctan if needed.
Time Calculation
The time it takes to cross the river is crucial for planning and safety in navigation. In this problem, the time depends on how fast you move perpendicular to the river's flow. Here, the boat moves east at 4.2 m/s. The formula is:
  • \[ t = \frac{\text{distance across}}{\text{velocity}_{perpendicular}} \]
For a width of 500 m, the calculation is straightforward. It allows you to determine how long it will take to reach the other bank, assuming a straight path relative to your own direction.
Drift Calculation
Drift is the distance you are carried downstream by the river's current during your journey across. Knowing how far you will drift helps in aligning your starting point with your intended landing spot. In this context, it is the southern movement due to the river's speed. The drift depends on the river's velocity and the crossing time. Use the equation:
  • \[ d = \text{velocity}_{downstream} \times \text{time}_{crossing} \]
This provides you a clear expectation of where you'll end up on the opposite bank. It's essential for plotting a successful and precise crossing.

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Most popular questions from this chapter

A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle \(\textbf{(Fig. P3.63). }\) The takeoff ramp was inclined at 53.0\(^{\circ}\), the river was 40.0 m wide, and the far bank was 15.0 m lower than the top of the ramp. The river itself was 100 m below the ramp. Ignore air resistance. (a) What should his speed have been at the top of the ramp to have just made it to the edge of the far bank? (b) If his speed was only half the value found in part (a), where did he land?

A major leaguer hits a baseball so that it leaves the bat at a speed of 30.0 m/s and at an angle of 36.9\(^\circ\) above the horizontal. Ignore air resistance. (a) At what \(two\) times is the baseball at a height of 10.0 m above the point at which it left the bat? (b) Calculate the horizontal and vertical components of the baseball's velocity at each of the two times calculated in part (a). (c) What are the magnitude and direction of the baseball's velocity when it returns to the level at which it left the bat?

Henrietta is jogging on the sidewalk at 3.05 m/s on the way to her physics class. Bruce realizes that she forgot her bag of bagels, so he runs to the window, which is 38.0 m above the street level and directly above the sidewalk, to throw the bag to her. He throws it horizontally 9.00 s after she has passed below the window, and she catches it on the run. Ignore air resistance. (a) With what initial speed must Bruce throw the bagels so that Henrietta can catch the bag just before it hits the ground? (b) Where is Henrietta when she catches the bagels?

A canoe has a velocity of 0.40 m/s southeast relative to the earth. The canoe is on a river that is flowing 0.50 m/s east relative to the earth. Find the velocity (magnitude and direction) of the canoe relative to the river.

A 76.0-kg rock is rolling horizontally at the top of a vertical cliff that is 20 m above the surface of a lake \(\textbf{(Fig. P3.65).}\) The top of the vertical face of a dam is located 100 m from the foot of the cliff, with the top of the dam level with the surface of the water in the lake. A level plain is 25 m below the top of the dam. (a) What must be the minimum speed of the rock just as it leaves the cliff so that it will reach the plain without striking the dam? (b) How far from the foot of the dam does the rock hit the plain?
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