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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 \(9.0 \mathrm{~km} / \mathrm{h}\) 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 km/h; (b) Upstream (positive x-axis); (c) 1 km/h; (d) Downstream (negative x-axis)

Step by step solution

01

Determine Boat's Velocity Relative to the Ground

To find the velocity of the boat with respect to the ground, we subtract the river's flow velocity from the boat's velocity because the boat is traveling upstream. The boat's velocity with respect to the water is \( 14 \text{ km/h}\), and the river flows at \( 9 \text{ km/h}\) in the opposite direction.\( v_{bg} = v_{bw} - v_w = 14 \text{ km/h} - 9 \text{ km/h} = 5 \text{ km/h} \).Thus, the magnitude of the boat's velocity with respect to the ground is \( 5 \text{ km/h} \).
02

Determine Direction of the Boat's Velocity

Since the boat is moving upstream against the current, the direction of the boat's velocity relative to the ground is upstream, in the positive direction of the \( x \) axis.
03

Find Child's Velocity with Respect to the Boat

The child walks from front to rear of the boat at \( 6 \text{ km/h} \) in the opposite direction to the boat's positive direction of travel.
04

Calculate Child's Velocity Relative to the Ground

The child's velocity with respect to the ground is a combination of their velocity with respect to the boat and the boat's velocity with respect to the ground. Since the child walks opposite the boat's direction, we subtract their speed from that of the boat:\( v_{cg} = v_{bg} - v_{cb} = 5 \text{ km/h} - 6 \text{ km/h} = -1 \text{ km/h} \).This means the child's velocity relative to the ground is \( 1 \text{ km/h} \) in the negative direction of the \( x \) axis.
05

Determine Direction of the Child's Velocity

Because the calculated velocity of the child relative to the ground is negative, the direction of the child's velocity with respect to the ground is downstream, or in the negative direction of the \( x \) axis.

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Key Concepts

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

Velocity of a Boat
When considering the velocity of a boat, we need to think about how fast the boat moves in relation to the water around it. This is known as the boat's velocity relative to the water. In the given problem, the boat travels upstream at 14 km/h against the water's current, which moves at 9 km/h. To determine how quickly and in which direction the boat moves relative to the ground, it's essential to account for the effect of the water's current.

When the boat travels upstream, it is moving opposite to the direction of the water current. Thus, to find the boat's velocity concerning the ground, we subtract the velocity of the water from the velocity of the boat itself: \( v_{bg} = v_{bw} - v_{w} \).

In this instance, it simplifies to \(14 \text{ km/h} - 9 \text{ km/h} = 5 \text{ km/h} \). This means the boat moves at 5 km/h relative to the ground, upstream, or in the positive direction of the x-axis.
Reference Frames
Understanding reference frames is crucial when dealing with velocities. A reference frame is essentially a point of view from which we measure and observe motion. For this problem, three different reference frames are considered:

  • The water as the reference frame for the boat. Here, the boat's velocity is 14 km/h.
  • The ground as the reference frame for both the water and the boat.
  • The boat itself as the reference frame for the child. The child is moving at 6 km/h within this frame.
By switching reference frames, we can determine velocities from different perspectives. When we compute the velocity of the boat with respect to the ground, we are switching from the water's frame to the ground's frame. This switch helps us see how the boat actually moves over solid ground, factoring in the opposing current from the river.
Motion in One Dimension
Motion in one dimension simplifies problems since movement occurs along a single straight line. In this context, both the boat's and the child's movements occur on the x-axis—marking it as one-dimensional motion.

Understanding this setup is crucial because it limits velocity to two possible directions: positive or negative. In the exercise, the boat moves upstream (positive direction), and the child walking in the opposite direction brings about a negative velocity component with respect to the boat.

The mathematics of motion in one dimension revolves around simple addition or subtraction of velocities, such as when calculating how the child's walking speed (opposing the boat's direction) influences their total velocity relative to ground. Thus, this form of motion simplifies the complexity often present in multi-dimensional motion problems.
Vector Addition
Vector addition is a method used to combine different velocities when they are in effect simultaneously. In our example, the boat's velocity with respect to the ground and the child's velocity demonstrate this concept.

Since the child walks in the opposite direction of the boat's movement, vector addition involves subtracting the child's walking speed from the boat's velocity relative to the ground. This concept applies as follows: \( v_{cg} = v_{bg} - v_{cb} \), where \(v_{cg}\) is the child's velocity with respect to the ground, \(v_{bg}\) is the boat's velocity relative to the ground, and \(v_{cb}\) is the child's velocity relative to the boat.

The result \(-1 \text{ km/h} \) denotes that the child's net velocity is directed downstream compared to the ground, highlighting the countereffect of vector addition when directions oppose each other.

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Most popular questions from this chapter

A train travels due south at \(30 \mathrm{~m} / \mathrm{s}\) (relative to the ground) in a rain that is blown toward the south by the wind. The path of each raindrop makes an angle of \(70^{\circ}\) with the vertical, as measured by an observer stationary on the ground. An observer on the train, however, sees the drops fall perfectly vertically. Determine the speed of the raindrops relative to the ground.

Long flights at midlatitudes in the Northern Hemisphere encounter the jet stream, an eastward airflow that can affect a plane's speed relative to Earth's surface. If a pilot maintains a certain speed relative to the air (the plane's airspeed), the speed relative to the surface (the plane's ground speed) is more when the flight is in the direction of the jet stream and less when the flight is opposite the jet stream. Suppose a round-trip flight is scheduled between two cities separated by \(4000 \mathrm{~km}\), with the outgoing flight in the direction of the jet stream and the return flight opposite it. The airline computer advises an airspeed of \(1000 \mathrm{~km} / \mathrm{h}\), for which the difference in flight times for the outgoing and return flights is \(70.0\) min. What jet-stream speed is the computer using?

The range of a projectile depends not only on \(v_{0}\) and \(\theta_{0}\) but also on the value \(g\) of the free-fall acceleration, which varies from place to place. In 1936 , Jesse Owens established a world's running broad jump record of \(8.09 \mathrm{~m}\) at the Olympic Games at Berlin (where \(g=9.8128 \mathrm{~m} / \mathrm{s}^{2}\) ). Assuming the same values of \(v_{0}\) and \(\theta_{0}\), by how much would his record have differed if he had competed instead in 1956 at Melbourne (where \(g=9.7999 \mathrm{~m} / \mathrm{s}^{2}\) )?

Oasis \(A\) is \(90 \mathrm{~km}\) due west of oasis \(B .\) A desert camel leaves \(A\) and takes \(50 \mathrm{~h}\) to walk \(75 \mathrm{~km}\) at \(37^{\circ}\) north of due east. Next it takes \(35 \mathrm{~h}\) to walk \(65 \mathrm{~km}\) due south. Then it rests for \(5.0 \mathrm{~h}\). What are the (a) magnitude and (b) direction of the camel's displacement relative to \(A\) at the resting point? From the time the camel leaves \(A\) until the end of the rest period, what are the (c) magnitude and (d) direction of its average velocity and (e) its average speed? The camel's last drink was at \(A ;\) it must be at \(B\) no more than \(120 \mathrm{~h}\) later for its next drink. If it is to reach \(B\) just in time, what must be the (f) magnitude and (g) direction of its average velocity after the rest period?

A ball is shot from the ground into the air. At a height of \(9.1 \mathrm{~m}\), its velocity is \(\vec{v}=(7.6 \hat{\mathrm{i}}+6.1 \mathrm{j}) \mathrm{m} / \mathrm{s}\), with \(\hat{\mathrm{i}}\) horizontal and upward. (a) To what maximum height does the ball rise? (b) What total horizontal distance does the ball travel? What are the (c) magnitude and (d) angle (below the horizontal) of the ball's velocity just before it hits the ground?

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