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Chapter 3: Problem 10

The magnitude of vector \(\vec{A}\) is \(35.0\) units and points in the direction \(325^{\circ}\) counterclockwise from the positive \(x\) -axis. Calculate the \(x\) -and \(y\) -components of this vector.

Short Answer

Expert verified
The \(x\)-component of the vector is given by: \(35.0 \times cos(325^{\circ})\). The \(y\)-component of the vector is given by: \(35.0 \times sin(325^{\circ})\).

Step by step solution

01

Understand what to calculate

The goal is to find the \(x\) and \(y\) components of the vector \(\vec{A}\). The problem provides the magnitude, or length, of vector \(\vec{A}\) which is \(35.0\) units and it provides the orientation, or direction, of \(\vec{A}\) which is \(325^{\circ}\) counterclockwise from the positive \(x\) -axis.
02

Determine the x-component of the vector

The formula to calculate the x-component of \(\vec{A}\) is \(A_x = A \times cos(\theta)\), where \(A\) is the magnitude of the vector and \(\theta\) is the angle. Plugging in the given values, we have \(A_x = 35.0 \times cos(325^{\circ})\). Compute this expression to find the x-component.
03

Determine the y-component of the vector

Similarly, the formula for the y-component of \(\vec{A}\) is \(A_y = A \times sin(\theta)\). Plugging in the given values, we have \(A_y = 35.0 \times sin(325^{\circ})\). Compute this to find the y-component.

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Key Concepts

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

Understanding Magnitude
Magnitude refers to the size or length of a vector. In this case, the magnitude of vector \(\vec{A}\) is given as \(35.0\) units.

Think of it like the distance a person walks from their house, irrespective of the direction.
It's a measure of how much of the vector there is. Calculating magnitude doesn't involve direction and is always a positive number.

When breaking down vectors into their components, magnitude is essential because it helps to determine the vector鈥檚 contribution in each direction.
It forms the hypotenuse of the right triangle created by the vector's components in a standard Cartesian coordinate system.
Angle Measurement in Vectors
Angle measurement is crucial in determining the direction of a vector.

In our example, the angle for vector \(\vec{A}\) is \(325^{\circ}\), measured counterclockwise from the positive x-axis.
This measurement helps in understanding where the vector is pointing. Angles are often measured in degrees, but they can also be expressed in radians.

To find the components of a vector, you'll use this angle to determine how much the vector points horizontally and vertically:
  • Horizontal component (x-component)
  • Vertical component (y-component)
By using trigonometric functions like cosine and sine, you can convert this angle into useful x and y values.
Applying Trigonometry to Vectors
Trigonometry is a mathematical tool that connects angles to ratios of side lengths in right triangles.

When working with vectors, trigonometric functions such as cosine and sine are key to finding components.
For a vector \(\vec{A}\) with magnitude \(A\) and angle \(\theta\):
  • The x-component \(A_x\) can be found using \(A \times \cos(\theta)\)
  • The y-component \(A_y\) can be found using \(A \times \sin(\theta)\)
In our exercise, you'd substitute \(35.0\) for \(A\) and \(325^{\circ}\) for \(\theta\), then calculate:
  • \(A_x = 35.0 \times \cos(325^{\circ})\)
  • \(A_y = 35.0 \times \sin(325^{\circ})\)
This approach allows you to break down any vector into simpler, manageable parts.

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Most popular questions from this chapter

A chinook (king) salmon (genus Oncorhynchus) can jump out of water with a speed of \(6.26 \mathrm{~m} / \mathrm{s}\). (See Problem \(4.9\), page 111 for an investigation of how the fish can leave the water at a higher speed than it can swim underwater.) If the salmon is in a stream with water speed equal to \(1.50 \mathrm{~m} / \mathrm{s}\), how high in the air can the fish jump if it leaves the water traveling vertically upwards relative to the Earth?

A \(2.00-\mathrm{m}\) -tall basketball player is standing on the floor \(10.0 \mathrm{~m}\) from the basket, as in Figure \(\mathrm{P} 3.58\). If he shoots the ball at a \(40.0^{\circ}\) angle with the horizontal, at what initial speed must he throw the basketball so that it goes through the hoop without striking the backboard? The height of the basket is \(3.05 \mathrm{~m}\).

In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slides off the counter and strikes the floor at distance \(d\) from the base of the counter. If the height of the counter is \(h\), (a) find an expression for the time \(t\) it takes the cup to fall to the floor in terms of the variables \(h\) and \(g\). (b) With what speed does the mug leave the counter? Answer in terms of the variables \(d, g\), and \(h .(c)\) In the same terms, what is the speed of the cup immediately before it hits the floor? (d) In terms of \(h\) and \(d\), what is the direction of the cup's velocity immediately before it hits the floor?

You can use any coordinate system you like to solve a projectile motion problem. To demonstrate the truth of this statement, consider a ball thrown off the top of a building with a velocity \(\vec{v}\) at an angle \(\theta\) with respect to the horizontal. Let the building be \(50.0 \mathrm{~m}\) tall, the initial horizontal velocity be \(9.00 \mathrm{~m} / \mathrm{s}\), and the initial vertical velocity be \(12.0 \mathrm{~m} / \mathrm{s}\). Choose your coordinates such that the positive \(y\) -axis is upward, the \(x\) -axis is to the right, and the origin is at the point where the ball is released. (a) With these choices, find the ball's maximum height above the ground and the time it takes to reach the maximum height. (b) Repeat your calculations choosing the origin at the base of the building.

A jet airliner moving initially at \(3.00 \times 10^{2} \mathrm{mi} / \mathrm{h}\) due east enters a region where the wind is blowing \(1.00\) \(\times 10^{2} \mathrm{mi} / \mathrm{h}\) in a direction \(30.0^{\circ}\) north of east. (a) Find the components of the velocity of the jet airliner relative to the air, \(\vec{v}_{\mu}\). (b) Find the components of the velocity of the air relative to Earth, \(\overrightarrow{\mathbf{v}}_{M \mathcal{E}}\). (c) Write an equation analo- gous to Equation \(3.16\) for the velocities \(\vec{v}_{j}, \vec{v}_{A B}\), and \(\vec{v}_{g E}\) (d) What is the speed and direction of the aircraft relative to the ground?
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