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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 \(8.2 \mathrm{~km} / \mathrm{h}\) 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) 5.8 km/h upstream, (b) upstream, (c) 0.2 km/h, (d) downstream

Step by step solution

01

Understand the Problem Statement

The problem deals with finding velocities of a boat and a child with respect to the ground. It involves the addition of vectors, as velocities in different frames of reference are given. Our task is to find out the combined effects, i.e., the resultant velocity.
02

Define the Known Quantities

Let's list what we know:- The boat's velocity with respect to the water: \(v_{bw} = 14 \, \text{km/h}\) (upstream direction)- The water's velocity with respect to the ground: \(v_{wg} = 8.2 \, \text{km/h}\) (downstream direction)- The child's velocity with respect to the boat: \(v_{cb} = -6.0 \, \text{km/h}\) (since walking from front to rear)
03

Calculate the Boat's Velocity with Respect to the Ground

The boat travels upstream against the flowing river. Thus, its velocity with respect to the ground is the difference between its velocity relative to the water and the river's speed:\[ v_{bg} = v_{bw} - v_{wg} = 14 \, \text{km/h} - 8.2 \, \text{km/h} = 5.8 \, \text{km/h} \]This magnitude is also the direction upstream.
04

Calculate the Child's Velocity with Respect to the Ground

Since the child is moving with respect to the boat and the boat is moving with respect to the ground, we add the velocities as vectors:\[ v_{cg} = v_{cb} + v_{bg} = (-6.0 \, \text{km/h}) + 5.8 \, \text{km/h} = -0.2 \, \text{km/h} \]The negative sign indicates that child moves in the downstream direction at \(0.2 \text{ km/h}\).

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Key Concepts

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

Vector Addition
When dealing with velocities, it's important to understand how vector addition plays a crucial role. In physics, a vector is a quantity that has both magnitude (how much) and direction (which way). Unlike numbers, vectors require special rules to add them up. This is because vectors can point in different directions, so you can't just add their magnitudes like ordinary numbers.

When you add vectors, you're essentially looking for a resultant vector that has a particular size and direction. Here's how it's commonly done:
  • For velocities in the same direction, you add their magnitudes.
  • For velocities in opposite directions, you subtract their magnitudes.
This rule is applicable in the given exercise where the boat's velocity relative to the water and the river's flow have to be subtracted because they move in opposite directions. By applying vector addition, we found the boat's velocity with respect to the ground. Remember, this is also how velocities between multiple moving objects, like the child on the boat, can be calculated.
Frame of Reference
The concept of a "frame of reference" helps us understand how motion is observed from different viewpoints. It is essentially the 'background' or 'perspective' from which we measure or observe motion. Think of a frame of reference like a camera that follows an object: the view changes based on the camera's position.

In our exercise, two frames of reference are given:
  • Water as a frame, from which we observe the boat moving upstream.
  • The ground as another frame, giving us the river's flow speed in that reference.
When we talk about "velocity with respect to the ground," we are viewing the movement from the ground's viewpoint. This is different from viewing movement from inside the boat or from the river's flow. Knowing which frame you're using is crucial as it can change the velocities you calculate.
Velocity with Respect to the Ground
Determining velocity with respect to the ground gives us insight into how fast an object is moving when observed from a stationary point on Earth. In the exercise, you calculate the boat's velocity by considering both the rowing speed and the river flow speed.

The formula for finding the velocity of an object relative to the ground can usually be written as:
\[ v_{ag} = v_{ab} + v_{bg} \]
where:
  • \(v_{ag}\) is the object's velocity with respect to the ground.
  • \(v_{ab}\) is the object's velocity with respect to a moving frame (like the boat).
  • \(v_{bg}\) is the velocity of the moving frame with respect to the ground (like the river flow).
In our example, we discovered this by determining how the boat's speed relative to water adjusted with river speed, and similarly for the child moving on the boat. The crucial part here is understanding how each relative motion combines to give a full picture of the true motion with respect to the ground.

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Most popular questions from this chapter

Music is still available on vinyl records that are played on turntables. Such a record rotates with a period of \(1.8 \mathrm{~s}\). For a record with a radius of \(16 \mathrm{~cm}\), find the centripetal accceleration of a point on the edge of the record.

Two ships, \(A\) and \(B\), leave port at the same time. Ship \(A\) travels northwest at 24 knots, and ship \(B\) travels at 28 knots in a direction \(40^{\circ}\) west of south \((1\) knot \(=1\) nautical mile per hour; see Appendix D). What are the (a) magnitude and (b) direction of the velocity of ship \(A\) relative to \(B\) ? (c) After what time will the ships be 160 nautical miles apart? (d) What will be the bearing of \(B\) (the direction of \(B\) 's position) relative to \(A\) at that time?

A cameraman on a pickup truck is traveling westward at \(20 \mathrm{~km} / \mathrm{h}\) while he records a cheetah that is moving westward \(30 \mathrm{~km} / \mathrm{h}\) faster than the truck. Suddenly, the cheetah stops, turns, and then runs at \(45 \mathrm{~km} / \mathrm{h}\) eastward, as measured by a suddenly nervous crew member who stands alongside the cheetah's path. The change in the animal's velocity takes \(2.0 \mathrm{~s}\). What are the (a) magnitude and (b) direction of the animal's acceleration according to the cameraman and the (c) magnitude and (d) direction according to the nervous crew member?

At \(t_{1}=2.00 \mathrm{~s}\), the acceleration of a particle in counterclockwise circular motion is \(\left(6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(4.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). It moves at constant speed. At time \(t_{2}=5.00 \mathrm{~s}\), the particle's acceleration is \((4.00\) \(\left.\mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(-6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). What is the radius of the path taken by the particle if \(t_{2}-t_{1}\) is less than one period?

A train at a constant \(60.0 \mathrm{~km} / \mathrm{h}\) moves east for \(40.0 \mathrm{~min}\), then in a direction \(50.0^{\circ}\) east of due north for \(20.0 \mathrm{~min}\), and then west for \(50.0 \mathrm{~min}\). What are the (a) magnitude and (b) angle of its average velocity during this trip?

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