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If the magnitude of a velocity vector is \(7.0 \mathrm{~m} / \mathrm{s}\) and the \(x\) -component is \(3.0 \mathrm{~m} / \mathrm{s},\) what is the \(y\) -component?

Short Answer

Expert verified
The y-component is approximately 6.3 m/s.

Step by step solution

01

Understand the Problem

The problem requires us to find the y-component of a velocity vector when the magnitude and the x-component are known. The velocity vector can be represented in terms of its components as \( v = \sqrt{v_x^2 + v_y^2} \).
02

Write Down the Known Values

From the problem, we know the magnitude of the velocity vector \( v = 7.0 \; \text{m/s} \) and the x-component \( v_x = 3.0 \; \text{m/s} \).
03

Set Up the Equation

Use the Pythagorean theorem for the components of the velocity vector: \( v = \sqrt{v_x^2 + v_y^2} \). Substitute the known values: \( 7.0 = \sqrt{(3.0)^2 + v_y^2} \).
04

Solve for the y-component

Square both sides of the equation to eliminate the square root: \( (7.0)^2 = (3.0)^2 + v_y^2 \). This gives \( 49 = 9 + v_y^2 \).
05

Rearrange to find the y-component

Rearrange the equation to solve for \( v_y^2 \): \( v_y^2 = 49 - 9 \). This simplifies to \( v_y^2 = 40 \).
06

Calculate the y-component

Take the square root of both sides to find \( v_y \): \( v_y = \sqrt{40} \). When you calculate it, \( v_y \approx 6.3 \; \text{m/s} \).

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Key Concepts

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

Velocity vector
In physics, a velocity vector represents the speed and direction of an object's motion. The vector allows us to describe both how fast something is moving and in which direction. By combining both magnitude (speed) and direction, it presents a much clearer picture of an object's motion than speed alone could.
For instance, if a car travels at 7 meters per second heading east, its velocity is a vector quantity, while its speed would just be 7 meters per second. To fully describe velocity, we use vector components, typically noted by their directional designation such as x-component and y-component, to show an object's movement in two dimensions.
Pythagorean theorem
The Pythagorean theorem is a fundamental principle in geometry that applies to right-angled triangles. It states: in a right 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 lengths of the other two sides.
This is represented as:
  • \( c^2 = a^2 + b^2 \)
where \( c \) is the hypotenuse, and \( a \) and \( b \) are the other two sides.
In the context of physics and velocity vectors, the theorem helps us determine the magnitude of the velocity vector using its components. When given the x and y components of a velocity, the Pythagorean theorem aids in finding the overall velocity magnitude, allowing us to compute unknown components. This makes it an invaluable tool in solving vector-related physics problems.
Vector components
Vector components break down vectors into their directional parts. In two-dimensional space, these typically involve an x-component (horizontal) and a y-component (vertical). This separation helps simplify complex vector operations and allows for easy calculation of individual elements.
When a vector is known, but only some of its individual components are given, we often use techniques like the Pythagorean theorem to solve for the missing parts. Vector components are crucial because they allow us to apply linear methods to what could otherwise be complex, non-linear problems, especially when dealing with different directions of motion, like in the velocity vector problem.
Physics equations
Physics equations, such as those used for calculating vectors, create a framework for understanding and predicting physical phenomena. These equations often leverage mathematical principles to describe the relationships between different physical quantities.
In the case of velocity vectors, the equation involving the Pythagorean theorem, \( v = \sqrt{v_x^2 + v_y^2} \), is used to relate the magnitude of the velocity with its x and y components. Each component acts as a building block, helping us to find other properties of the system being analyzed. By rearranging these equations, we can isolate unknowns to find solutions, such as determining the missing y-component when other values are known.

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Most popular questions from this chapter

During part of its trajectory (which lasts exactly \(1 \mathrm{~min}\) ) a missile travels at a constant speed of \(2000 \mathrm{mi} / \mathrm{h}\) while maintaining a constant orientation angle of \(20^{\circ}\) from the vertical. (a) During this phase, what is true about its velocity components: \((1) v_{y}>v_{x},\) (2) \(v_{y}=v_{x},\) or (3) \(v_{y}

A boat that travels at a speed of \(6.75 \mathrm{~m} / \mathrm{s}\) in still water is to go directly across a river and back ( \(>\) Fig. 3.38 ). The current flows at \(0.50 \mathrm{~m} / \mathrm{s}\). (a) At what angle(s) must the boat be steered? (b) How long does it take to make the round trip? (Assume that the boat's speed is constant at all times, and neglect turnaround time.)

A ball rolling on a table has a velocity with rectangular components \(v_{x}=0.60 \mathrm{~m} / \mathrm{s}\) and \(v_{y}=0.80 \mathrm{~m} / \mathrm{s} .\) What is the displacement of the ball in an interval of \(2.5 \mathrm{~s} ?\)

An airplane is flying at \(150 \mathrm{mi} / \mathrm{h}\) (its speed in still air) in a direction such that with a wind of \(60.0 \mathrm{mi} / \mathrm{h}\) blowing from east to west, the airplane travels in a straight line southward. (a) What must be the plane's heading (direction) for it to fly directly south? (b) If the plane has to go \(200 \mathrm{mi}\) in the southward direction, how long does it take?

A ball is projected horizontally with an initial speed of \(5.0 \mathrm{~m} / \mathrm{s}\). Find its (a) position and (b) velocity at \(t=2.5 \mathrm{~s}\)

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