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Chapter 3: Problem 2

A rhinoceros 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 \(t_{2}=12.0 \mathrm{~s},\) (a) what are the \(x\) - and \(y\) -coordinates of the rhino? (b) How far is the rhino from the origin?

Short Answer

Expert verified
a) The x-coordinate is -45.6m and the y-coordinate is 58.8m. b) The distance from the origin is 74.13m.

Step by step solution

01

Calculate the x-coordinate

The average \(x\)-velocity is \(-3.8 \mathrm{~m} / \mathrm{s}\), and the total time is \(12.0 \mathrm{~s}\). The distance in the \(x\)-direction, which is also the \(x\)-coordinate, can be calculated using the formula \(d = v \cdot t = -3.8 \mathrm{~m/s} \cdot 12.0 \mathrm{s} = -45.6 \mathrm{m}\).
02

Calculate the y-coordinate

Similarly, the average \(y\)-velocity is \(4.9 \mathrm{~m} / \mathrm{s}\), and the total time remains \(12.0 \mathrm{~s}\). The distance in the \(y\)-direction, which is also the \(y\)-coordinate, can be calculated using the formula \(d = v \cdot t = 4.9 \mathrm{~m/s} \cdot 12.0 \mathrm{s} = 58.8 \mathrm{m}\).
03

Calculate the distance from the origin

Now to calculate the distance of the rhino from the origin, we use the Pythagorean theorem. This can be expressed as \(\sqrt{x^2 + y^2}\). Using the numbers calculated from the previous steps gives \(\sqrt{(-45.6 m)^2 + (58.8 m)^2} = 74.13 \mathrm{m}\).

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Key Concepts

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

average velocity
In kinematics, average velocity is a vector quantity that represents the total displacement divided by the total time taken. It encompasses both magnitude and direction, unlike speed, which is scalar. Average velocity is crucial in determining how fast and in which direction an object is moving over a certain period.

To understand average velocity, consider it as the change in position (displacement) over time, represented mathematically as:
  • Average Velocity, \( \vec{v}_{avg} = \frac{\Delta \vec{d}}{\Delta t} \), where \( \Delta \vec{d} \) is displacement and \( \Delta t \) is the time interval.
  • In vector form: \( \vec{v}_{avg} = \langle v_{x}, v_{y} \rangle \), where \( v_x \) and \( v_y \) are components along the x and y axes.
Given this, the average x-velocity of the rhino is \(-3.8 \text{ m/s}\) and the y-velocity is \(4.9 \text{ m/s}\), signifying not just how fast, but also in which directions the rhino is consistently moving over the 12 seconds.
coordinate systems
Coordinate systems serve as frameworks for locating points in space. In the two-dimensional plane, typically represented as the XY-plane, every point is positioned by an x-coordinate and a y-coordinate.

For a problem like the rhino's movement:
  • The x-coordinate indicates the horizontal position.
  • The y-coordinate specifies the vertical position.
At the start, the rhino is at the origin \((0,0)\). With the average velocities known, you can calculate the final coordinates.
  • Use the average velocity in each direction and multiply by the time to find the displacement in that direction:
  • For the x-coordinate: the rhino moves \(-45.6 \text{ m}\).
  • For the y-coordinate: the rhino displaces \(58.8 \text{ m}\).
So, the final position after 12 seconds of movement is \((-45.6, 58.8)\). Imagine tracing this path in your mind to visualize where the rhino ends up in the coordinate system.
distance calculation
Finding the distance a moving object is from its starting position involves understanding a bit of geometry. This is where the Pythagorean theorem becomes helpful.

When you have both x and y displacements, you can think of the total distance as a line connecting the origin to the final point, forming a right triangle:
  • The legs are the x and y displacements.
  • The hypotenuse is the direct distance from the start to the end point.
The Pythagorean theorem is expressed as:
  • \( d = \sqrt{x^2 + y^2} \)
  • For the rhino: substituting the new coordinates gives \( \sqrt{(-45.6)^2 + (58.8)^2} = 74.13 \text{ m} \).
This formula calculates the straight-line or Euclidean distance from the starting point to the rhino's final location. This clear path-finding aids in understanding both the magnitude of travel and the directness of a journey path.

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Most popular questions from this chapter

In fighting forest fires, airplanes work in support of ground crews by dropping water on the fires. For practice, a pilot drops a canister of red dye, hoping to hit a target on the ground below. If the plane is flying in a horizontal path \(90.0 \mathrm{~m}\) above the ground and has a speed of \(64.0 \mathrm{~m} / \mathrm{s}(143 \mathrm{mi} / \mathrm{h}),\) at what horizontal distance from the target should the pilot release the canister? Ignore air resistance.

The radius of the earth's orbit around the sun (assumed to be circular) is \(1.50 \times 10^{8} \mathrm{~km},\) and the earth travels around this orbit in 365 days. (a) What is the magnitude of the orbital velocity of the earth, in \(\mathrm{m} / \mathrm{s} ?(\mathrm{~b})\) What is the magnitude of the radial acceleration of the earth toward the sun, in \(\mathrm{m} / \mathrm{s}^{2} ?\) (c) Repeat parts (a) and (b) for the motion of the planet Mercury (orbit radius \(=5.79 \times 10^{7} \mathrm{~km},\) orbital period \(=88.0\) days).

A firefighting crew uses a water cannon that shoots water at \(25.0 \mathrm{~m} / \mathrm{s}\) at a fixed angle of \(53.0^{\circ}\) above the horizontal. The firefighters want to direct the water at a blaze that is \(10.0 \mathrm{~m}\) above ground level. How far from the building should they position their cannon? There are \(t w o\) possibilities; can you get them both? (Hint: Start with a sketch showing the trajectory of the water.

A \(76.0 \mathrm{~kg}\) rock is rolling horizontally at the top of a vertical cliff that is \(20 \mathrm{~m}\) above the surface of a lake (Fig. \(\mathrm{P} 3.65\) ). The top of the vertical face of a dam is located \(100 \mathrm{~m}\) from the foot of the cliff, with the top of the dam level with the surface of the water in the lake. A level plain is \(25 \mathrm{~m}\) below the top of the dam. (a) What must be the minimum speed of the rock just as it leaves the cliff so that it will reach the plain without striking the dam? (b) How far from the foot of the dam does the rock hit the plain?

In a World Cup soccer match, Juan is running due north toward the goal with a speed of \(8.00 \mathrm{~m} / \mathrm{s}\) relative to the ground. A teammate passes the ball to him. The ball has a speed of \(12.0 \mathrm{~m} / \mathrm{s}\) and is moving in a direction \(37.0^{\circ}\) east of north, relative to the ground. What are the magnitude and direction of the ball's velocity relative to Juan?
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