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The velocity of a particle in reference frame \(A\) is \((2.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} .\) The velocity of reference frame \(A\) with respect to reference frame \(B\) is \(4.0 \hat{\mathbf{k}} \mathrm{m} / \mathrm{s}, \quad\) and the velocity of reference frame \(B\) with respect to \(C\) is \(2.0 \mathrm{j} \mathrm{m} / \mathrm{s} .\) What is the velocity of the particle in reference frame \(C\) ?

Short Answer

Expert verified
The velocity of the particle in Reference Frame C is \(\vec{v}_{p/C} = 2.0\hat{\mathbf{i}} + 5.0\hat{\mathbf{j}} + 4.0\hat{\mathbf{k}}\) m/s.

Step by step solution

01

Write down the given information

We are given: - Velocity of the particle in Reference Frame A: \(\vec{v}_{p/A} = 2.0\hat{\mathbf{i}} + 3.0\hat{\mathbf{j}}\) m/s - Velocity of Reference Frame A with respect to Reference Frame B: \(\vec{v}_{A/B} = 4.0\hat{\mathbf{k}}\) m/s - Velocity of Reference Frame B with respect to Reference Frame C: \(\vec{v}_{B/C} = 2.0\hat{\mathbf{j}}\) m/s
02

Apply relative velocity formula

In order to find the velocity of the particle in Reference Frame C, \(\vec{v}_{p/C}\), we need to apply the relative velocity formula: \(\vec{v}_{p/C} = \vec{v}_{p/A} + \vec{v}_{A/C}\) However, we need to convert the velocity of Reference Frame A with respect to Reference Frame B, and the velocity of Reference Frame B with respect to Reference Frame C, to the velocity of Reference Frame A with respect to Reference Frame C: \(\vec{v}_{A/C} = \vec{v}_{A/B} + \vec{v}_{B/C}\)
03

Calculate the velocity of Reference Frame A with respect to Reference Frame C

Using the given values, calculate \(\vec{v}_{A/C}\): \(\vec{v}_{A/C} = (4.0\hat{\mathbf{k}}) + (2.0\hat{\mathbf{j}}) = 2.0\hat{\mathbf{j}} + 4.0\hat{\mathbf{k}}\)
04

Calculate the velocity of the particle in Reference Frame C

Now, plug the calculated value \(\vec{v}_{A/C}\) into the relative velocity formula to determine the velocity of the particle in Reference Frame C: \(\vec{v}_{p/C} = (2.0\hat{\mathbf{i}} + 3.0\hat{\mathbf{j}}) + (2.0\hat{\mathbf{j}} + 4.0\hat{\mathbf{k}}) = 2.0\hat{\mathbf{i}} + 5.0\hat{\mathbf{j}} + 4.0\hat{\mathbf{k}}\) So, the velocity of the particle in Reference Frame C is: \(\vec{v}_{p/C} = 2.0\hat{\mathbf{i}} + 5.0\hat{\mathbf{j}} + 4.0\hat{\mathbf{k}}\) m/s.

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Key Concepts

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

Reference Frame
Imagine you're standing still watching a car drive past you, and then picture yourself inside another car watching the same vehicle go by. Your perspective in each scenario differs, right? This is where the term reference frame comes into play. In physics, a reference frame is a perspective from which an observer measures and describes the motion of objects. It's like choosing a unique vantage point to observe events unfolding around you.

There are two types of reference frames: inertial and non-inertial. An inertial reference frame is one that is either at rest or moves with a constant velocity, meaning it's not accelerating. On the other hand, a non-inertial reference frame is accelerating, making motion appear differently due to forces like gravity or friction. In our exercise, the problem specifies different reference frames (A, B, and C), and understanding how they relate to each other is crucial for analyzing the motion of the particle.
Velocity Vector
A velocity vector represents an object's speed and direction—like an arrow flying toward a target, its length illustrates how fast, and the arrow's point shows where it's heading. In our exercise, the velocity vectors are denoted in a three-dimensional coordinate system using unit vectors \( \hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}} \) for the x, y, and z-directions, respectively.

For example, the particle's velocity in reference frame A is expressed as \( \vec{v}_{p/A} = 2.0\hat{\mathbf{i}} + 3.0\hat{\mathbf{j}} \) m/s, indicating a combined horizontal and vertical movement but no motion along the z-axis. Comprehending how these vectors work is essential to dealing with motion in multiple dimensions.
Relative Motion
The concept of relative motion stems from the idea that movement can look different depending on where you're observing from. For instance, if you're standing on a conveyor belt, you feel like you're moving even though you might be still relative to someone walking alongside the belt. Similarly, in physics, we use relative motion to describe how the velocity of an object appears to change when observed from various reference frames.

In our exercise, we're tasked with determining the particle's velocity relative to reference frame C, so we need to consider how frames A and B are moving relative to C. Combining these motions allows us to find the particle's true velocity from C's point of view, which can be radically different from how it would appear from A or B.
Vector Addition
To understand how velocities combine in different reference frames, we use vector addition, the method of adding two or more vectors together. You can visualize it by placing the tail of one arrow against the head of the other, forming a 'vector chain.' The resulting arrow, stretching from the starting point to the final tip, represents the combined effect of these vectors.

In the context of the exercise, we added the velocities of reference frames A and B using vector addition to find A's velocity concerning C, showing \( \vec{v}_{A/C} = 4.0\hat{\mathbf{k}} + 2.0\hat{\mathbf{j}} \) m/s. We then used the same method to sum the particle's velocity in frame A with A's velocity relative to frame C to determine the particle's velocity in frame C, yielding \( \vec{v}_{p/C} = 2.0\hat{\mathbf{i}} + 5.0\hat{\mathbf{j}} + 4.0\hat{\mathbf{k}} \) m/s. Thus, vector addition is key to analyzing complex movements in physics.

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Most popular questions from this chapter

You throw a baseball at an initial speed of \(15.0 \mathrm{m} / \mathrm{s}\) at an angle of \(30^{\circ}\) with respect to the horizontal. What would the ball's initial speed have to be at \(30^{\circ}\) on a planet that has twice the acceleration of gravity as Earth to achieve the same range? Consider launch and impact on a horizontal surface.

Clay Matthews, a linebacker for the Green Bay Packers, can reach a speed of \(10.0 \mathrm{m} / \mathrm{s}\). At the start of a play, Matthews runs downfield at \(45^{\circ}\) with respect to the 50-yard line and covers \(8.0 \mathrm{m}\) in \(1 \mathrm{s}\). He then runs straight down the field at \(90^{\circ}\) with respect to the 50 -yard line for \(12 \mathrm{m},\) with an elapsed time of \(1.2 \mathrm{s} .\) (a) What is Matthews' final displacement from the start of the play? (b) What is his average velocity?

A particle is executing circular motion with a constant angular frequency of \(\omega=4.00 \mathrm{rad} / \mathrm{s} .\) If time \(t=0\) corresponds to the position of the particle being located at \(y=0 \mathrm{m}\) and \(x=5 \mathrm{m},\) (a) what is the position of the particle at \(t=10 \mathrm{s} ?\) (b) What is its velocity at this time? (c) What is its acceleration?

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An airplane flying horizontally with a speed of 500 \(\mathrm{km} / \mathrm{h}\) at a height of \(800 \mathrm{m}\) drops a crate of supplies (see the following figure). If the parachute fails to open, how far in front of the release point does the crate hit the ground?

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