You want to cross a straight section of a river that has a uniform current of \(5.33 \mathrm{~m} / \mathrm{s}\) and is \(127, \mathrm{~m}\) wide. Your motorboat has an engine that can generate a speed of \(17.5 \mathrm{~m} / \mathrm{s}\) for your boat. Assume that you reach top speed right away (that is, neglect the time it takes to accelerate the boat to top speed). a) If you want to go directly across the river with a \(90^{\circ}\) angle relative to the riverbank, at what angle relative to the riverbank should you point your boat? b) How long will it take to cross the river in this way? c) In which direction should you aim your boat to achieve minimum crossing time? d) What is the minimum time to cross the river? e) What is the minimum speed of your boat that will still enable you to cross the river with a \(90^{\circ}\) angle relative to the riverbank?

Short Answer

Expert verified
Answer: The boat should maintain an angle of approximately 17.42 degrees to go directly across the river. It takes roughly 8.24 seconds to cross at this angle. The minimum time to cross the river is about 7.26 seconds. The minimum speed needed to maintain a 90-degree angle relative to the riverbank is 5.33 m/s.

Step by step solution

01

Calculate the angle to go directly across the river

To find the angle at which the boat should be pointed, we can use the equation tan(θ) = v_current / v_an, where v_current is the velocity of the current, v_an is the boat's velocity perpendicular to the current, and θ is the angle we want to find. Solving for θ, we get: θ = arctan(v_current / v_an) Here, v_current = 5.33 m/s and v_an = sqrt(17.5^2 - 5.33^2) m/s (since the boat's speed is the resultant of the boat's speed perpendicular to the river current and speed along the river current). θ = arctan(5.33 / sqrt(17.5^2 - 5.33^2)) ≈ 17.42 degrees
02

Calculate the time taken to cross the river

We can now find the time taken to cross the river by using the formula t = d / v_an, where d is the width of the river and v_an is the velocity of the boat perpendicular to the river current. The width of the river, d = 127 m, and v_an = sqrt(17.5^2 - 5.33^2) m/s. t = 127 / sqrt(17.5^2 - 5.33^2) ≈ 8.24 seconds
03

Determine the direction for minimum crossing time

To achieve the minimum crossing time, the boat should be aimed directly towards the opposite bank, perpendicular to the river current. This means the boat should be pointed at a 90-degree angle relative to the riverbank.
04

Calculate the minimum time to cross the river

Using the same formula as in step 2, t = d / v_an, but this time, the boat is pointed directly across the river (90-degree angle), so v_an = 17.5 m/s. t_min = 127 / 17.5 ≈ 7.26 seconds
05

Calculate the minimum speed needed to maintain a 90-degree angle

To maintain a 90-degree angle relative to the riverbank, the boat's velocity should be equal to the river current's velocity. Therefore, the minimum speed needed to cross the river at a 90-degree angle is 5.33 m/s.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An object moves in the \(x y\) -plane. The \(x\) - and \(y\) -coordinates of the object as a function of time are given by the following equations: \(x(t)=4.9 t^{2}+2 t+1\) and \(y(t)=3 t+2 .\) What is the velocity vector of the object as a function of time? What is its acceleration vector at a time \(t=2\) s?

Two cannonballs are shot from different cannons at angles \(\theta_{01}=20^{\circ}\) and \(\theta_{02}=30^{\circ}\), respectively. Assuming ideal projectile motion, the ratio of the launching speeds, \(v_{01} / v_{02},\) for which the two cannonballs achieve the same range is a) \(0.742 \mathrm{~m}\) d) \(1.093 \mathrm{~m}\) b) \(0.862 \mathrm{~m}\) e) \(2.222 \mathrm{~m}\) c) \(1.212 \mathrm{~m}\)

In ideal projectile motion, when the positive \(y\) -axis is chosen to be vertically upward, the \(y\) -component of the acceleration of the object during the ascending part of the motion and the \(y\) -component of the acceleration during the descending part of the motion are, respectively, a) positive, negative. c) positive, positive. b) negative, positive. d) negative, negative.

Neglect air resistance for the following. A soccer ball is kicked from the ground into the air. When the ball is at a height of \(12.5 \mathrm{~m},\) its velocity is \((5.6 \hat{x}+4.1 \hat{y}) \mathrm{m} / \mathrm{s}\). a) To what maximum height will the ball rise? b) What horizontal distance will be traveled by the ball? c) With what velocity (magnitude and direction) will it hit the ground?

An arrow is shot horizontally with a speed of \(20 \mathrm{~m} / \mathrm{s}\) from the top of a tower \(60 \mathrm{~m}\) high. The time to reach the ground will be a) \(8.9 \mathrm{~s}\) c) \(3.5 \mathrm{~s}\) e) \(1.0 \mathrm{~s}\) b) \(7.1 \mathrm{~s}\) d) \(2.6 \mathrm{~s}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free