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Chapter 3: Problem 42

A jogger runs with a speed of \(3.25 \mathrm{m} / \mathrm{s}\) in a direction \(30.0^{\circ}\) above the \(x\) axis. (a) Find the \(x\) and \(y\) components of the jogger's velocity. (b) How will the velocity components found in part (a) change if the jogger 's speed is halved?

Short Answer

Expert verified
a) \(v_x = 2.82 \, \mathrm{m/s}\), \(v_y = 1.625 \, \mathrm{m/s}\); b) \(v'_x = 1.41 \, \mathrm{m/s}\), \(v'_y = 0.8125 \, \mathrm{m/s}\).

Step by step solution

01

Understanding Velocity Components

To find the components of velocity, use the trigonometric functions cosine and sine related to the angle given. The cosine function will give us the horizontal component (\(x\)-component), and the sine function will yield the vertical component (\(y\)-component). The general formula is: \[ v_x = v\cos(\theta) \] \[ v_y = v\sin(\theta) \] where \(v\) is the speed and \(\theta\) is the angle.
02

Calculate the x-component of Velocity

Apply the formula for the \(x\)-component: \[ v_x = 3.25\cos(30.0^{\circ}) \] Knowing that \(\cos(30.0^{\circ}) = \frac{\sqrt{3}}{2} \approx 0.866 \), we find: \[ v_x = 3.25 \times 0.866 \approx 2.82 \, \mathrm{m/s} \]
03

Calculate the y-component of Velocity

Use the formula for the \(y\)-component: \[ v_y = 3.25\sin(30.0^{\circ}) \] Knowing that \(\sin(30.0^{\circ}) = 0.5\), we find: \[ v_y = 3.25 \times 0.5 = 1.625 \, \mathrm{m/s} \]
04

Assess the Impact of Halved Speed

If the jogger's speed is halved, the new speed is \(\frac{3.25}{2} = 1.625 \, \mathrm{m/s}\). Apply the same trigonometric functions with the halved speed to find the new components.
05

Calculate New x-component with Halved Speed

With the halved speed, the \(x\)-component becomes: \[ v'_x = 1.625\cos(30.0^{\circ}) \approx 1.625 \times 0.866 = 1.41 \, \mathrm{m/s} \]
06

Calculate New y-component with Halved Speed

With the halved speed, the \(y\)-component becomes: \[ v'_y = 1.625\sin(30.0^{\circ}) = 1.625 \times 0.5 = 0.8125 \, \mathrm{m/s} \]

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Key Concepts

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

trigonometric functions
Trigonometric functions are crucial in physics for breaking down vectors into components, especially when dealing with angles. When you have a vector at an angle, it doesn't simply translate into a straight-line movement. Instead, it has two parts: a horizontal component and a vertical component.
By using trigonometric functions such as **cosine** and **sine**, we can determine these components easily:
  • The cosine function (\(\cos(\theta)\)) helps to find the horizontal, or \(x\)-component.
  • The sine function (\(\sin(\theta)\)) is used to find the vertical, or \(y\)-component.
The general formulas to find these components are:
  • \(v_x = v\cos(\theta)\)
  • \(v_y = v\sin(\theta)\)
This allows us to deconstruct any vector into understandable parts using basic trigonometry.
velocity components
Velocity components describe different aspects of motion for an object moving at an angle. A single velocity vector can be split into horizontal and vertical components. This decomposition makes it simpler to analyze the motion through specific directions.
For a jogger running at a speed of 3.25 m/s and an angle of 30 degrees above the x-axis:
  • The horizontal component (\(v_x\)) indicates how fast the jogger is moving along the x-axis: \(v_x = 3.25\times0.866 \approx 2.82 \, \mathrm{m/s}\).
  • The vertical component (\(v_y\)) reflects the upward or downward motion: \(v_y = 3.25\times0.5 = 1.625 \, \mathrm{m/s}\).
By understanding both components, we have a clearer picture of how the jogger moves through space.
kinematic equations
Kinematic equations often describe motion in terms of velocity, acceleration, time, and displacement. When analyzing motion, especially in two dimensions, it's essential to break it into manageable parts.
In this scenario, when the jogger's speed is halved (new speed: 1.625 m/s), both velocity components must be recalculated:
  • For the new horizontal component: \(v'_x = 1.625\times0.866 \approx 1.41 \, \mathrm{m/s}\).
  • For the new vertical component: \(v'_y = 1.625\times0.5 = 0.8125 \, \mathrm{m/s}\).
These calculations show that reducing the speed affects each component individually, maintaining their proportions but decreasing their magnitudes. Understanding these basics of kinematic equations helps to breakdown and solve more complex motion problems effectively.

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Most popular questions from this chapter

As you hurry to catch your flight at the local airport, you encounter a moving walkway that is \(85 \mathrm{m}\) long and has a speed of \(2.2 \mathrm{m} / \mathrm{s}\) relative to the ground. If it takes you \(68 \mathrm{s}\) to cover \(85 \mathrm{m}\) when walking on the ground, how long will it take you to cover the same distance on the walkway? Assume that you walk with the same speed on the walkway as you do on the ground.

A passenger walks from one side of a ferry to the other as it approaches a dock. If the passenger's velocity is \(1.50 \mathrm{m} / \mathrm{s}\) due north relative to the ferry, and \(4.50 \mathrm{m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) west of north relative to the water, what are the direction and magnitude of the ferry's velocity relative to the water?

A One-Percent Grade A road that rises 1 ft for every \(100 \mathrm{ft}\) traveled horizontally is said to have a \(1 \%\) grade. Portions of the Lewiston grade, near Lewiston, Idaho, have a \(6 \%\) grade. At what angle is this road inclined above the horizontal?

As a function of time, the velocity of the football described in Problem 68 can be written as \(\overrightarrow{\mathbf{v}}=(16.6 \mathrm{m} / \mathrm{s}) \hat{\mathrm{x}}-\left[\left(9.81 \mathrm{m} / \mathrm{s}^{2}\right) t\right] \hat{\mathrm{y}}\) Calculate the average acceleration vector of the football for the time periods (a) \(t=0\) to \(t=1.00 \mathrm{s},\) (b) \(t=0\) to \(t=2.50 \mathrm{s},\) and (c) \(t=0\) to \(t=5.00 \mathrm{s}\). (If the acceleration of an object is constant, its average acceleration is the same for all time periods.)

A Big Clock The clock that rings the bell known as Big Ben has an hour hand that is 9.0 feet long and a minute hand that is 14 feet long, where the distance is measured from the center of the clock to the tip of each hand. What is the tip-to-tip distance between these two hands when the clock reads 12 minutes after four o'clock?
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