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Chapter 3: Problem 56

Relative to the ground, a car has a velocity of \(18.0 \mathrm{~m} / \mathrm{s},\) directed due north. Relative to this car, a truck has a velocity of \(22.8 \mathrm{~m} / \mathrm{s}\), directed \(52.1^{\circ}\) south of east. Find the magnitude and direction of the truck's velocity relative to the ground.

Short Answer

Expert verified
The truck's velocity relative to the ground is approximately 22.3 m/s, at an angle of 35.4掳 north of east.

Step by step solution

01

Define Velocities

Let the car's velocity relative to the ground be denoted as \( \vec{v}_{cg} \), and it is given as \( 18.0 \mathrm{~m/s} \) due north. Let the truck's velocity relative to the car be \( \vec{v}_{tc} = 22.8 \mathrm{~m/s} \) at \( 52.1^{\circ} \) south of east.
02

Break Down the Truck's Velocity into Components

We will break down \( \vec{v}_{tc} = 22.8 \mathrm{~m/s} \) into its east (\( v_{tcx} \)) and south (\( v_{tcy} \)) components: \( v_{tcx} = 22.8 \cos(52.1^{\circ}) \) and \( v_{tcy} = 22.8 \sin(52.1^{\circ}) \).
03

Apply Vector Addition to Find Truck's Velocity Relative to Ground

The truck's velocity relative to the ground \( \vec{v}_{tg} \) can be determined by adding the vectors \( \vec{v}_{cg} \) and \( \vec{v}_{tc} \): \[ \vec{v}_{tg} = \vec{v}_{cg} + \vec{v}_{tc} \]. This involves adding the x-components and y-components separately.
04

Calculate the x and y Components

The x-component of \( \vec{v}_{tg} \) in the east direction is \( v_{tgx} = v_{tcx} = 22.8 \cos(52.1^{\circ}) \). The y-component of \( \vec{v}_{tg} \) in the north direction is \( v_{tgy} = v_{cg} - v_{tcy} = 18.0 - 22.8 \sin(52.1^{\circ}) \).
05

Find the Magnitude of the Truck's Velocity Relative to the Ground

Calculate the magnitude of \( \vec{v}_{tg} \) using the Pythagorean theorem: \[ |\vec{v}_{tg}| = \sqrt{(v_{tgx})^2 + (v_{tgy})^2} \].
06

Determine the Direction of the Truck's Velocity Relative to the Ground

Calculate the direction angle \( \theta \) using the tangent inverse (arctan) function: \[ \theta = \tan^{-1}\left(\frac{v_{tgy}}{v_{tgx}}\right) \]. The direction is measured from the east toward the north, making adjustments if needed based on component signs.

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Key Concepts

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

Vector Addition
Vector addition is a crucial concept in physics, especially when analyzing motion like calculating relative velocities.
Vectors contain both magnitude and direction, for example, how fast something is moving (magnitude) and in which direction (north, east, etc.).
To add vectors, you need to combine their respective components.
  • The component along the x-axis gives the east-west direction.
  • The component along the y-axis gives the north-south direction.
In the problem, the goal is to find the velocity of the truck relative to the ground by adding the velocity of the car relative to the ground and truck relative to the car by aligning their vector components correctly.
Velocity Components
Breaking down a vector into its components is essential when dealing with directions that are not aligned with the principal axes (north, south, east, west). This process helps simplify vector calculations.
For the truck's velocity, we split it into two perpendicular directions:
  • The east direction (x-component): Found using the cosine of the angle.
  • The south direction (y-component): Found using the sine of the angle.
The formulas used are:
  • East component: \( v_{tcx} = 22.8 \cos(52.1^{\circ}) \)
  • South component: \( v_{tcy} = 22.8 \sin(52.1^{\circ}) \)
These components are used in combination with the car's velocity to find the composite velocity vector relative to the ground.
Pythagorean Theorem
The Pythagorean theorem is a powerful tool to find the magnitude of a resultant vector, which is a combination of its x and y components. It's based on the relationship in right triangles: \[ c^2 = a^2 + b^2 \] where \( c \) is the hypotenuse.
In vector analysis, the hypotenuse represents the overall magnitude of the vector.
To find the truck's velocity relative to the ground:
  • First, you find the x and y components as discussed.
  • Apply the Pythagorean theorem with these components.
The calculation is: \[ |\vec{v}_{tg}| = \sqrt{(v_{tgx})^2 + (v_{tgy})^2} \] which gives the truck's speed relative to the ground.
Direction Angle
Once you have the magnitude, determining the direction is crucial for fully understanding the vector's effect. The direction angle gives us this orientation, expressed relative to a standard axis (east, north, etc.).
Using trigonometry, specifically the tangent function, helps find this angle.
The formula \( \theta = \tan^{-1}\left(\frac{v_{tgy}}{v_{tgx}}\right) \) allows the calculation of the angle formed between the resultant velocity vector and the east axis.
  • Make adjustments based on the signs of the components to ensure the angle reflects the correct quadrant of the direction.
  • In this scenario, the east and slightly north direction must be determined for proper navigation and understanding.
Understanding how to find the direction angle helps interpret the resultant velocity in real-world terms such as sailing or flying directions.

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Most popular questions from this chapter

A jetliner is moving at a speed of \(245 \mathrm{~m} / \mathrm{s}\). The vertical component of the plane's velocity is \(40.6 \mathrm{~m} / \mathrm{s}\). Determine the magnitude of the horizontal component of the plane's velocity.

A rifle is used to shoot twice at a target, using identical cartridges. The first time, the rifle is aimed parallel to the ground and directly at the center of the bull's-eye. The bullet strikes the target at a distance of \(H_{\mathrm{A}}\) below the center, however. The second time, the rifle is similarly aimed, but from twice the distance from the target. This time the bullet strikes the target at a distance of \(H_{\mathrm{B}}\) below the center. Find the ratio \(H_{\mathrm{B}} / H_{\mathrm{A}}\).

A radar antenna is tracking a satellite orbiting the earth. At a certain time, the radar screen shows the satellite to be \(162 \mathrm{~km}\) away. The radar antenna is pointing upward at an angle of \(62.3^{\circ}\) from the ground. Find the \(x\) and \(y\) components (in \(\mathrm{km}\) ) of the position of the satellite.

(a) When a projectile is launched horizontally from a rooftop at a speed \(v_{0 x}\), does its horizontal velocity component ever change in the absence of air resistance? (b) Can you calculate the horizontal distance \(D\) traveled after launch simply as \(D=v_{0 x} t\), where \(t\) is the fall time of the projectile? (c) In calculating the fall time, is the vertical part of the motion just like that of a ball dropped from rest? Explain each of your answers. A criminal is escaping across a rooftop and runs off the roof horizontally at a speed of \(5.3 \mathrm{~m} / \mathrm{s}\), hoping to land on the roof of an adjacent building. The horizontal distance between the two buildings is \(D,\) and the roof of the adjacent building is \(2.0 \mathrm{~m}\) below the jumping- off point. Find the maximum value for \(D\).

A Coast Guard ship is traveling at a constant velocity of \(4.20 \mathrm{~m} / \mathrm{s},\) due east, relative to the water. On his radar screen the navigator detects an object that is moving at a constant velocity. The object is located at a distance of \(2310 \mathrm{~m}\) with respect to the ship, in a direction \(32.0^{\circ}\) south of east. Six minutes later, he notes that the object's position relative to the ship has changed to \(1120 \mathrm{~m}, 57.0^{\circ}\) south of west. What are the magnitude and direction of the velocity of the object relative to the water? Express the direction as an angle with respect to due west.
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