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Chapter 3: Problem 5

The \(x\) -component of a velocity vector that has an angle of \(37^{\circ}\) to the \(+x\) -axis has a magnitude of \(4.8 \mathrm{~m} / \mathrm{s}\) (a) What is the magnitude of the velocity? (b) What is the magnitude of the \(y\) -component of the velocity?

Short Answer

Expert verified
(a) 6.01 m/s (b) 3.62 m/s.

Step by step solution

01

Understanding the given information

We are given the x-component of the velocity vector and its angle with the +x-axis. The x-component of the velocity is \( v_x = 4.8 \text{ m/s} \), and the angle \( \theta = 37^\circ \). We need to find the magnitude of the entire velocity vector \( v \) and the y-component of the velocity \( v_y \).
02

Calculating the magnitude of the velocity vector

The magnitude of the velocity vector can be found using the formula \( v_x = v \cdot \cos(\theta) \). To find \( v \), we rearrange the formula to get \( v = \frac{v_x}{\cos(\theta)} \). Substitute \( v_x = 4.8 \text{ m/s} \) and \( \cos(37^\circ) \approx 0.7986 \) into the equation: \[v = \frac{4.8}{0.7986} \approx 6.01 \text{ m/s}.\]
03

Calculating the y-component of the velocity

The y-component of the velocity vector can be found using \( v_y = v \cdot \sin(\theta) \). Substitute the previously calculated \( v = 6.01 \text{ m/s} \) and \( \sin(37^\circ) \approx 0.6018 \) into the equation: \[v_y = 6.01 \cdot 0.6018 \approx 3.62 \text{ m/s}.\]

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Key Concepts

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

Understanding Velocity Calculation
Velocity plays a crucial role in describing motion. In physics, velocity is not just the speed an object is moving but also the direction in which it moves. When calculating velocity, particularly for a vector at an angle, it often helps to break it into its component parts.
To calculate the velocity magnitude from its horizontal component, we use trigonometric relationships. The formula we're interested in is:
  • \(v_x = v \cdot \cos(\theta)\),
    where \(v_x\) is the given x-component of the velocity, \(v\) is the magnitude of the entire velocity vector, and \(\theta\) is the angle with the x-axis.
Rearranging helps us find \(v\), the magnitude of the velocity:\[v = \frac{v_x}{\cos(\theta)}\].
This equation allows us to calculate the full magnitude of the velocity when we know one component and the angle with respect to the x-axis.
Using Trigonometry in Physics
Trigonometry is a powerful tool in physics, particularly when dealing with vectors like velocity. It helps relate the angles and sides of triangles to solve problems involving directions and magnitudes. In this scenario, we apply trigonometry to resolve a velocity vector into its components.
We employ two main trigonometric functions here:
  • Cosine Function: Used to determine the x-component of a vector. For our problem:
    \(v_x = v \cdot \cos(37^{\circ})\).
  • Sine Function: Used for finding the y-component:
    \(v_y = v \cdot \sin(37^{\circ})\).
These functions help break down intricate vector problems into simpler calculations. Calculating the components separately makes it easier to analyze motion along individual axes before considering the overall movement.
Finding the Magnitude of Velocity
The magnitude of velocity gives us the total speed of the object irrespective of its direction.
After understanding and using the x-component, we derive the magnitude of the entire velocity. Starting with:\[v = \frac{4.8}{0.7986} \approx 6.01 \text{ m/s}\],
we find the total velocity to be approximately 6.01 m/s. This scalar tells us how fast the object is moving in space overall, distinct from its x and y movements.For the y-component, we again turn to trigonometry:\[v_y = 6.01 \times 0.6018 \approx 3.62 \text{ m/s}\].
This calculation gives us insight into how much of the total velocity is directed along the y-axis, helping visualize not just speed but also how an object transitions through space.

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Most popular questions from this chapter

An observer by the side of a straight, level, northsouth road watches a car (A) moving south at a rate of \(75 \mathrm{~km} / \mathrm{h}\). A driver in another car (B) going north at \(50 \mathrm{~km} / \mathrm{h}\) also observes car \(\mathrm{A}\). (a) What is car A's velocity as observed from car B? (Take north to be positive.) (b) If the roadside observer sees car A brake to a stop in \(6.0 \mathrm{~s}\) what constant acceleration would be measured? (c) What constant acceleration would the driver in car B measure for the braking car A?

A ball rolls horizontally with a speed of \(7.6 \mathrm{~m} / \mathrm{s}\) off the edge of a tall platform. If the ball lands \(8.7 \mathrm{~m}\) from the point on the ground directly below the edge of the platform, what is the height of the platform?

In laboratory situations, a projectile's range can be used to determine its speed. To see how this is done, suppose a ball rolls off a horizontal table and lands \(1.5 \mathrm{~m}\) out from the edge of the table. If the tabletop is \(90 \mathrm{~cm}\) above the floor, determine (a) the time the ball is in the air, and (b) the ball's speed as it left the table top.

A golfer lines up for her first putt at a hole that is \(10.5 \mathrm{~m}\) exactly northwest of her ball's location. She hits the ball \(10.5 \mathrm{~m}\) and straight, but at the wrong angle, \(40^{\circ}\) from due north. In order for the golfer to have a "twoputt green," determine (a) the angle of the second putt and (b) the magnitude of the second putt's displacement. (c) Determine why you cannot determine the length of travel of the second putt.

(a) If each of the two components \((x\) and \(y\) ) of a vector are doubled, (1) the vector's magnitude doubles, but the direction remains unchanged; (2) the vector's magnitude remains unchanged, but the direction angle doubles; or (3) both the vector's magnitude and direction angle double. (b) If the \(x\) - and \(y\) -components of a vector of \(10 \mathrm{~m}\) at \(45^{\circ}\) are tripled, what is the new vector?
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