/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 70 A boat is traveling upstream in ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A boat is traveling upstream in the positive direction of an \(x\) axis at \(14 \mathrm{~km} / \mathrm{h}\) with respect to the water of a river. The water is flowing at \(9.0 \mathrm{~km} / \mathrm{h}\) with respect to the ground. What are the (a) magnitude and (b) direction of the boat's velocity with respect to the ground? A child on the boat walks from front to rear at \(6.0 \mathrm{~km} / \mathrm{h}\) with respect to the boat. What are the (c) magnitude and (d) direction of the child's velocity with respect to the ground?

Short Answer

Expert verified
(a) 5.0 km/h upstream, (c) 1.0 km/h downstream.

Step by step solution

01

Analyze the Boat's Velocity Relative to the Ground

The boat is moving upstream at \(14 \, \text{km/h}\) with respect to the water and the water itself is flowing at \(9.0 \, \text{km/h}\) with respect to the ground. Since the boat is moving upstream against the current, the velocity of the boat with respect to the ground \(v_{bg}\) is given by the equation:\[ v_{bg} = v_{bw} - v_{wg} = 14 \, \text{km/h} - 9.0 \, \text{km/h} = 5.0 \, \text{km/h} \]
02

Determine the Boat's Direction Relative to the Ground

The boat's velocity with respect to the ground is positive because it is moving upstream, which is defined as the positive direction for this problem. Therefore, the direction of the boat's velocity with respect to the ground is upstream.
03

Analyze the Child's Velocity Relative to the Ground

The child is walking from the front to the rear of the boat at \(6.0 \, \text{km/h}\) with respect to the boat, which is moving upstream. Considering the boat's movement, the child's velocity \(v_{cg}\) with respect to the ground is given by:\[ v_{cg} = v_{cb} + v_{bg} = -6.0 \, \text{km/h} + 5.0 \, \text{km/h} = -1.0 \, \text{km/h} \] The negative sign indicates that the velocity direction is opposite to the positive direction.
04

Determine the Child's Direction Relative to the Ground

The child's velocity with respect to the ground is \(-1.0 \, \text{km/h}\), indicating motion in the negative direction. This means the child is moving downstream with respect to the ground.

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 91Ó°ÊÓ!

Key Concepts

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

Upstream and Downstream
When we discuss objects moving in water, like a boat, terms like upstream and downstream come into play. These terms help describe the relative direction of movement concerning the water's current.
If a boat moves upstream, it is going against the direction of water flow. Downstream means moving along with the current. For example, if a river flows north and a boat travels south, it is going upstream. Conversely, moving northward with the flow is traveling downstream.
Understanding these directions is crucial because they affect the calculations of velocity in relation to other objects, like when a boat moves in a current.
Velocity with Respect to Ground
When determining an object's velocity with respect to the ground, we need to consider both its speed and its direction. This concept is especially important when objects, like boats or people, are in a moving medium such as water or air.
For example, a boat moving upstream would have its velocity decreased by the current's speed. This is because you subtract the current's velocity from the boat's velocity. Conversely, when moving downstream, you add the current's velocity to the boat's velocity, resulting in a faster speed with respect to the ground.
  • To find an object's velocity with respect to the ground, consider the combined effect of its movement and the flow of its surrounding medium.
  • Use the formula for relative velocity: \[ v_{bg} = v_{bw} - v_{wg} \] for upstream and \[ v_{bg} = v_{bw} + v_{wg} \] for downstream.
Vector Addition
Vector addition is a mathematical operation that allows us to combine different velocities or forces. It’s crucial for solving problems involving relative velocities.
Imagine you’re facing both upstream and downstream movements, like with a boat in a river. You need to combine these different components to determine the actual velocity in relation to a fixed point, like the ground.
  • Vectors are quantities that have both magnitude and direction, like velocity.
  • Adding vectors involves considering both of these components: \[ v_{cg} = v_{cb} + v_{bg} \]for calculating the child's velocity with respect to the ground. Here, each part of the vector equation considers the different relative motions.
This calculation helps provide clarity on complex movements by breaking them down into simpler, additive parts.

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

A rugby player runs with the ball directly toward his opponent's goal, along the positive direction of an \(x\) axis. He can legally pass the ball to a teammate as long as the ball's velocity relative to the field does not have a positive \(x\) component. Suppose the player runs at speed \(4.0 \mathrm{~m} / \mathrm{s}\) relative to the field while he passes the ball with velocity \(\vec{v}_{B P}\) relative to himself. If \(\vec{v}_{B P}\) has magnitude \(6.0 \mathrm{~m} / \mathrm{s}\), what is the smallest angle it can have for the pass to be legal?

A ball is thrown horizontally from a height of \(20 \mathrm{~m}\) and hits the ground with a speed that is three times its initial speed. What is the initial speed?

A particle moves along a circular path over a horizontal \(x y\) coordinate system, at constant speed. At time \(t_{1}=4.00 \mathrm{~s}\), it is at point \((5.00 \mathrm{~m}, 6.00 \mathrm{~m})\) with velocity \((3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) and acceleration in the positive \(x\) direction. At time \(t_{2}=10.0 \mathrm{~s}\), it has velocity \((-3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}\) and acceleration in the positive \(y\) direction. What are the (a) \(x\) and (b) \(y\) coordinates of the center of the circular path if \(t_{2}-t_{1}\) is less than one period?

A baseball is hit at ground level. The ball reaches its maximum height above ground level \(3.0 \mathrm{~s}\) after being hit. Then \(2.5 \mathrm{~s}\) after reaching its maximum height, the ball barely clears a fence that is \(97.5 \mathrm{~m}\) from where it was hit. Assume the ground is level. (a) What maximum height above ground level is reached by the ball? (b) How high is the fence? (c) How far beyond the fence does the ball strike the ground?

A car travels around a flat circle on the ground, at a constant speed of \(12.0 \mathrm{~m} / \mathrm{s}\). At a certain instant the car has an acceleration of \(3.00 \mathrm{~m} / \mathrm{s}^{2}\) toward the east. What are its distance and direction from the center of the circle at that instant if it is traveling (a) clockwise around the circle and (b) counterclockwise around the circle?

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.