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Relative to the ground, a car has a velocity of 16.0 m/s, directed due north. Relative to this car, a truck has a velocity of 24.0 m/s, directed 52.0 north of east. What is the magnitude of the truck’s velocity relative to the ground?

Short Answer

Expert verified
The truck's velocity relative to the ground is approximately 37.9 m/s.

Step by step solution

01

Understanding the Problem

We are given two velocity vectors: The car's velocity relative to the ground (\( \mathbf{v}_{cg} = 16 \, \text{m/s} \, \text{due north} \)) and the truck's velocity relative to the car (\( \mathbf{v}_{tc} = 24 \, \text{m/s} \, \text{at} \, 52^\circ \, \text{north of east} \)). The task is to find the truck's velocity relative to the ground (\( \mathbf{v}_{tg} \)).
02

Breaking Down Velocity of Truck Relative to Car

To find \( \mathbf{v}_{tg} \), we first resolve \( \mathbf{v}_{tc} \) into its components. Given its magnitude of 24.0 m/s and a direction of 52.0° north of east:\[ v_{tcx} = 24.0 \, \cos(52^\circ) \]\[ v_{tcy} = 24.0 \, \sin(52^\circ) \]
03

Calculating Components

Now, we calculate:\[ v_{tcx} = 24.0 \, \cos(52^\circ) \approx 14.8 \, \text{m/s} \] \[ v_{tcy} = 24.0 \, \sin(52^\circ) \approx 18.9 \, \text{m/s} \]
04

Determine Truck's Velocity Relative to Ground

To find \( \mathbf{v}_{tg} \), the x-component is directly \( v_{tcx} \) since the northward component of the car's motion doesn't affect it; for the y-component, the truck's northward component needs to be added to the car's:\[ v_{tgx} = v_{tcx} = 14.8 \, \text{m/s} \]\[ v_{tgy} = v_{cg} + v_{tcy} = 16.0 \, \text{m/s} + 18.9 \, \text{m/s} = 34.9 \, \text{m/s} \]
05

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

Now, combine the components using Pythagorean theorem to find the magnitude:\[ v_{tg} = \sqrt{(v_{tgx})^2 + (v_{tgy})^2} \]\[ v_{tg} = \sqrt{(14.8)^2 + (34.9)^2} \approx \sqrt{219.04 + 1218.01} \approx \sqrt{1437.05} \approx 37.9 \, \text{m/s} \]

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Key Concepts

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

Vector Addition
When studying physics, especially relative velocity, vector addition becomes an essential tool. Imagine vectors as arrows pointing in different directions. Each vector has both magnitude (how long it is) and direction (where it points).

In the exercise we're tackling, two vectors are involved: the car's velocity relative to the ground and the truck's velocity relative to the car. To find the truck's velocity relative to the ground, you add these vectors together. It's like piecing together a puzzle to see the bigger picture.

Key points to remember about vector addition in this context include:
  • Vectors can be broken down into components—think of it as splitting the arrow into smaller arrows that align with the x (horizontal) and y (vertical) axes.
  • To add vectors, add their respective components: x with x, and y with y.
  • The result is a new vector that combines influences of both original vectors.
This is how we transform multiple forces into a single vector that clearly shows overall impact.
Trigonometry in Physics
Trigonometry often pops up in physics, especially when dealing with angles and vectors. It gives us mathematical tools to resolve vectors into their components—those smaller arrows lying along the x and y axes.

In the problem at hand, the truck's velocity is at an angle, 52 degrees "north of east". To break it down, we use trigonometric functions like sine and cosine:

* Cosine helps us find the component of a vector along the x-axis (horizontal direction).
* Sine is used for the y-axis (vertical direction) component.

The formulas go like this:
  • Horizontal component: magnitude * cos(angle)
  • Vertical component: magnitude * sin(angle)
This way, the original directed vector splits into understandable parts, ready to be analyzed further. Mastering trigonometry in physics helps in facing a wide array of real-world scenarios.
Pythagorean Theorem
The Pythagorean theorem is a fundamental principle used in geometry and physics. It helps measure the length of the hypotenuse in a right triangle. When combined with vector components, it provides a powerful method to calculate overall magnitudes.

In relative velocity problems, after breaking vectors into x and y components, we often need the magnitude of the resulting vector. That's where Pythagorean theorem comes in. For a vector whose components you know, the theorem states:
  • \[ c = \sqrt{a^2 + b^2} \]
Here, \( c \) is the length of the hypotenuse (resultant vector), and \( a \) and \( b \) are the lengths of the other two sides (components).

In this lesson, when we calculated the truck's velocity relative to the ground, we used this theorem. After identifying and adding the components, the theorem provided the mush-needed magnitude, tying everything together in a neat solution.

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Most popular questions from this chapter

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