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A velocity vector has a magnitude of \(25.0 \mathrm{~m} / \mathrm{s}\). If its \(y\) component is \(-13.0 \mathrm{~m} / \mathrm{s},\) what are the possible values of its \(x\) component?

Short Answer

Expert verified
The possible values of the x-component are \(21.35\, \text{m/s}\) and \(-21.35\, \text{m/s}\).

Step by step solution

01

Understand the Problem

We are given a velocity vector with a known magnitude and a known y-component. We need to find the possible x-component values that satisfy these conditions.
02

Velocity Magnitude Formula

The magnitude of a velocity vector with components \(v_x\) and \(v_y\) is given by the formula: \( v = \sqrt{v_x^2 + v_y^2} \).
03

Substitute Known Values

We know the magnitude is \(25.0\, \text{m/s}\) and the \(y\) component is \(-13.0\, \text{m/s}\). Substitute these values into the formula: \( 25.0 = \sqrt{v_x^2 + (-13.0)^2} \).
04

Simplify the Equation

Calculate \((-13.0)^2\) to get \(169\). Substitute this into the equation: \( 25.0 = \sqrt{v_x^2 + 169} \). Then, square both sides to remove the square root: \( 625 = v_x^2 + 169 \).
05

Solve for \(v_x\)

Rearrange the equation to find \(v_x^2\): \( v_x^2 = 625 - 169 \). Calculate the result: \( v_x^2 = 456 \).
06

Calculate Possible Values of \(v_x\)

Take the square root of both sides: \( v_x = \pm \sqrt{456} \). Calculate \( \sqrt{456} \approx 21.35 \). Therefore, the possible values for \(v_x\) are approximately \(21.35\, \text{m/s}\) and \(-21.35\, \text{m/s}\).

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Key Concepts

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

Velocity Vector
A velocity vector represents both the speed and direction of an object in motion. This makes it different from speed, which is only a scalar quantity. A velocity vector consists of two main components in a two-dimensional space: the x-component and the y-component. These components are vectors themselves and point in the direction of motion along each axis of a coordinate system.
The x-component, denoted as \(v_x\), signifies how much of the velocity is in the horizontal direction. Meanwhile, the y-component, identified as \(v_y\), indicates the velocity in the vertical direction. Understanding these components helps in resolving the overall vector into simpler parts, which makes calculations, like determining an unknown component, more straightforward.
Magnitude Calculation
The magnitude of a velocity vector is a measure of the object's speed. It is calculated by finding the resultant of its components. In mathematical terms, this is determined using a formula that incorporates both the x-component and y-component of the vector. The standard formula to calculate the magnitude \(v\) of a velocity vector is:
  • \( v = \sqrt{v_x^2 + v_y^2} \)
This equation arises because the velocity vector forms a right triangle with its components. The square of the hypotenuse (magnitude) equals the sum of the squares of its other two sides. Applying this formula requires knowing or determining the individual components. In scenarios where one component and the overall magnitude are known, as in the problem provided, this formula allows us to find the missing component accurately.
Pythagorean Theorem
The Pythagorean theorem is a fundamental principle used extensively in geometry and physics to relate the sides of a right triangle. It states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This theorem is represented mathematically by the equation:
  • \( c^2 = a^2 + b^2 \)
Here, \(c\) is the hypotenuse, while \(a\) and \(b\) are the two other sides. This relationship is at the heart of the magnitude formula for vectors. When a velocity vector is decomposed into its x and y components, it forms a right triangle. Thus, by applying the Pythagorean theorem, one can determine the magnitude of the vector and solve for any missing component if the other data is provided.
Coordinate System
A coordinate system is a framework used to locate points in space. It helps us describe the position of objects using numbers, called coordinates. The most common system is the Cartesian coordinate system in two dimensions, made up of two perpendicular axes: the x-axis (horizontal) and the y-axis (vertical).
In physics and engineering, especially when dealing with vectors, understanding the coordinate system is crucial. The position and direction of vectors are defined concerning these axes. Vectors can shift direction and still maintain the same magnitude, simply by changing their components relative to the coordinate axes.
  • The position along the x-axis is given by the x-component \(v_x\).
  • The position along the y-axis is represented by the y-component \(v_y\).
Working within a coordinate system simplifies resolving combined motions into simpler parts and makes the analysis of such motions more intuitive and manageable.

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Most popular questions from this chapter

A test rocket is launched by accelerating it along a \(200.0-\mathrm{m}\) incline at \(1.25 \mathrm{~m} / \mathrm{s}^{2}\) starting from rest at point \(A\) (Figure 3.38 ). The incline rises at \(35.0^{\circ}\) above the horizontal, and at the instant the rocket leaves it, its engines turn off and it is subject only to gravity (air resistance can be ignored). Find (a) the maximum height above the ground that the rocket reaches, and (b) the greatest horizontal range of the rocket beyond point \(A\).

You swing a \(2.2 \mathrm{~kg}\) stone in a circle of radius \(75 \mathrm{~cm}\). At what speed should you swing it so its centripetal acceleration will be \(9.8 \mathrm{~m} / \mathrm{s}^{2} ?\)

A particle starts from rest at the origin with an acceleration vector that has magnitude \(4 \mathrm{~m} / \mathrm{s}^{2}\) and direction \(30^{\circ}\) above the positive \(x\) axis. (a) What are the components of its velocity vector \(20 \mathrm{~s}\) later? (b) What is the particle's position at that time?

A canoe has a velocity of \(0.40 \mathrm{~m} / \mathrm{s}\) southeast relative to the earth. The canoe is on a river that is flowing \(0.50 \mathrm{~m} / \mathrm{s}\) east relative to the earth. Find the velocity (magnitude and direction) of the canoe relative to the river.

A daring swimmer dives off a cliff with a running horizontal leap, as shown in Figure \(3.35 .\) What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is \(1.75 \mathrm{~m}\) wide and \(9.00 \mathrm{~m}\) below the top of the cliff?

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