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A rininoceros is at the origin of coordinates at time \(t_{1}=0 .\) For the time interval from \(t_{1}=0\) to \(t_{2}=12.0 \mathrm{s}\) , the rhino's average velocity has \(x-\) component \(-3.8 \mathrm{m} / \mathrm{s}\) and \(y\) -component 4.9 \(\mathrm{m} / \mathrm{s}\) . At time \(t_{2}=12.0 \mathrm{s},(\mathrm{a})\) what are the \(x\) - and \(y\) -coordinates of the rhino? (b) How far is the rhino from the origin?

Short Answer

Expert verified
At \(t = 12.0\, \text{s}\), the coordinates are \(-45.6 \, \text{m}, 58.8 \, \text{m}\), and distance from the origin is approximately \(74.35 \, \text{m}\).

Step by step solution

01

Find the x-coordinate

Use the formula for average velocity to calculate the x-coordinate. The average velocity in the x-direction is given by: \( v_x = \frac{\Delta x}{\Delta t} \), where \( \Delta x \) is the change in position and \( \Delta t \) is the time interval. Rearrange to find \( \Delta x = v_x \cdot \Delta t \). Given \( v_x = -3.8 \, \text{m/s} \) and \( \Delta t = 12.0 \, \text{s} \), compute \( \Delta x = -3.8 \, \text{m/s} \times 12.0 \, \text{s} = -45.6 \, \text{m} \). So, the x-coordinate at \( t = 12.0 \, \text{s} \) is \( -45.6 \, \text{m} \).
02

Find the y-coordinate

Use the formula for average velocity to calculate the y-coordinate. The average velocity in the y-direction is given by: \( v_y = \frac{\Delta y}{\Delta t} \). Rearrange to find \( \Delta y = v_y \cdot \Delta t \). Given \( v_y = 4.9 \, \text{m/s} \) and \( \Delta t = 12.0 \, \text{s} \), compute \( \Delta y = 4.9 \, \text{m/s} \times 12.0 \, \text{s} = 58.8 \, \text{m} \). So, the y-coordinate at \( t = 12.0 \, \text{s} \) is \( 58.8 \, \text{m} \).
03

Calculate the distance from the origin

The distance \( d \) from the origin is calculated using the Pythagorean theorem: \( d = \sqrt{x^2 + y^2} \). Substitute the x and y coordinates into this formula: \( d = \sqrt{(-45.6 \, \text{m})^2 + (58.8 \, \text{m})^2} \). Compute and find: \( d \approx \sqrt{2079.36 + 3451.84} \approx \sqrt{5531.2} \approx 74.35 \, \text{m} \).

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Key Concepts

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

Coordinate Calculation
In order to understand movement on a two-dimensional plane, we often use coordinate calculations. When dealing with problems involving motion, knowing how to calculate coordinates from average velocity is essential. If we’re tracking an object like a rhino from point to point over time, we first set a starting point (the origin) on a graph with an X and Y-axis corresponding to directions in our space.
To find the position of an object at any time, we calculate its change in position, known as displacement. The formula connecting average velocity and displacement in a given direction is crucial:
  • For the x-axis: \( \Delta x = v_x \times \Delta t \)
  • For the y-axis: \( \Delta y = v_y \times \Delta t \)
Using the provided components of average velocity, we can determine the exact position of the rhino after 12 seconds. These calculations result in x and y coordinates which solve part (a) of the original problem.
Pythagorean Theorem
The Pythagorean theorem is a fundamental concept in geometry, especially when dealing with right triangles. But it has significant applications in physics, particularly for problems involving distances in two-dimensional motion.
In a right triangle, the theorem states:
  • \( c = \sqrt{a^2 + b^2} \)
where \( c \) is the hypotenuse (the longest side) and \( a \) and \( b \) are the other two sides.
For our motion problem, the x and y displacements act as two perpendicular sides of a right triangle. Finding the straight-line distance from the origin to the rhino's position involves computing the hypotenuse of this triangle. Therefore, substituting the previously found x and y coordinates into the theorem gives us the distance from the origin.
Motion Problem
Motion problems often involve tracking an object as it moves over time, considering factors like speed, velocity, and displacement. In this exercise, the rhino starts at the origin, and its movement over a period is quantified using average velocity in different directions. Understanding how to apply average velocity is key, as it describes how fast and in what direction an object is moving on average over a time interval.
Practically, this means:
  • A negative x-component of velocity suggests motion to the left or backwards along the x-axis.
  • A positive y-component suggests upward or forward movement along the y-axis.
This exercise teaches us how to convert these concepts into changes in position over time and helps solve real-world physics problems.
Velocity Components
Velocity components are an essential concept when dealing with motion in two dimensions. In scenarios like the exercise, where an object moves slightly diagonally, it’s useful to break this motion into two perpendicular components—one for each axis.
The velocity vector can be dissected into:
  • \( v_x \): velocity in the horizontal direction
  • \( v_y \): velocity in the vertical direction
Understanding these components lets us calculate separate movements in each direction before combining them to find the total effect. In our problem, the rhino's average velocity is given with both x and y components, allowing for straightforward calculations of its position over time.

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Most popular questions from this chapter

Win the Prize. In a carnival booth, you win a stuffed giraffe if you toss a quarter into a small dish. The dish is on a shelf above the point where the quarter leaves your hand and is a horizontal distance of \(2.1 \mathrm{m}\) from this point (Fig. 3.41 ). If you toss the coin with a velocity of \(6.4 \mathrm{m} / \mathrm{s}\) at an angle of \(60^{\circ}\) above the horizontal, the coin lands in the dish. You can ignore air resistance. (a) What is the height of the shelf above the point where the quarter leaves your hand? (b) What is the vertical component of the velocity of the quarter just before it lands in the dish?

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In a test of a "g-suit," a volunteer is rotated in a horizontal circle of radius 7.0 \(\mathrm{m}\) . What must the period of rotation be so that the centripetal ncceleration has a magnitude of \((a) 3.0 g ?^{7}(b) 10 g ?\)

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