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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 \(t_{2}=12.0\) s, the rhino's aver- age 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
The coordinates are (-45.6 m, 58.8 m); distance from origin is 74.36 m.

Step by step solution

01

Understanding Average Velocity

Average velocity is defined as the displacement divided by the time interval. Here, the rhino's average velocity has components: \( v_{x_{avg}} = -3.8 \text{ m/s} \) and \( v_{y_{avg}} = 4.9 \text{ m/s} \).
02

Calculate Displacement in x-direction

Displacement in the \( x \)-direction \( (\Delta x) \) is given by \( \Delta x = v_{x_{avg}} \times \Delta t \). Substituting the values: \( \Delta x = -3.8 \text{ m/s} \times 12 \text{ s} = -45.6 \text{ m} \).
03

Calculate Displacement in y-direction

Displacement in the \( y \)-direction \( (\Delta y) \) is given by \( \Delta y = v_{y_{avg}} \times \Delta t \). Substituting the values: \( \Delta y = 4.9 \text{ m/s} \times 12 \text{ s} = 58.8 \text{ m} \).
04

Determine Coordinates at t=12s

Since the rhino starts at the origin, the final coordinates will be the displacement values in each direction. Thus, at \( t_2 = 12 \text{ s} \), the coordinates are \((x, y) = (-45.6 \text{ m}, 58.8 \text{ m})\).
05

Calculate Distance from Origin

The distance from the origin can be calculated using the Pythagorean theorem, \( d = \sqrt{x^2 + y^2} \). Substitute the values: \( d = \sqrt{(-45.6)^2 + (58.8)^2} = \sqrt{2079.36 + 3453.44} = \sqrt{5532.8} \approx 74.36 \text{ m} \).

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Key Concepts

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

Average Velocity
Average velocity is a central concept in kinematics. It is defined as the total displacement divided by the total time interval. In this exercise, the rhino's average velocity is provided with two components: an x-component of \(-3.8 \, \text{m/s}\) and a y-component of \(4.9 \, \text{m/s}\). These values tell us how much the rhino is moving in each direction per second.
To find the overall movement from these velocities, we multiply each component by the time interval. This gives us the displacement in both the x and y directions. Understanding average velocity helps in visualizing motion, as it directly connects the time spent moving and the resultant change in position.
Displacement Calculation
Displacement is the net change in position from the starting point to the end point during the time interval. For our rhino problem:
  • The displacement in the x-direction is calculated using the formula \(\Delta x = v_{x_{\text{avg}}} \times \Delta t\). Substituting the given values, we find \(\Delta x = -3.8 \, \text{m/s} \times 12 \, \text{s} = -45.6 \, \text{m}\).
  • Similarly, the y-direction displacement is \(\Delta y = v_{y_{\text{avg}}} \times \Delta t\), so \(\Delta y = 4.9 \, \text{m/s} \times 12 \, \text{s} = 58.8 \, \text{m}\).
These calculations show us the exact position changes in both horizontal and vertical directions. When the rhino starts at the origin, these displacements directly give the coordinates (-45.6 m, 58.8 m) after 12 seconds. This clear understanding of displacement is crucial for solving many other physics problems involving motion.
Pythagorean Theorem
The Pythagorean theorem is a mathematical relationship used to determine the distance between points in a plane. It states that, in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In the case of our rhino problem, we use this theorem to calculate the distance from the origin to the rhino's final position. The formula is \(d = \sqrt{x^2 + y^2}\).
  • Here, substitute \(x = -45.6 \text{ m}\) and \(y = 58.8 \text{ m}\).
  • Calculate to find \(d = \sqrt{(-45.6)^2 + (58.8)^2} = \sqrt{2079.36 + 3453.44} = \sqrt{5532.8} \approx 74.36 \text{ m}\).
Thus, the distance from the starting point to the endpoint is about 74.36 meters. This calculation not only answers the problem but also demonstrates the power of combining geometry with kinematics to understand spatial motions.

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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. E3.19). 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?

A web page designer creates an animation in which a dot on a computer screen has a position of \(\vec{r}=[4.0 \mathrm{cm}+\) \(\left(2.5 \mathrm{cm} / \mathrm{s}^{2}\right) t^{2} ] \hat{\boldsymbol{\imath}}+(5.0 \mathrm{cm} / \mathrm{s}) t \hat{\boldsymbol{J}}\) (a) Find the magnitude and direction of the dot's average velocity between \(t=0\) and \(t=2.0 \mathrm{s} .\) (b) Find the magnitude and direction of the instantaneous velocity at \(t=0, t=1.0 \mathrm{s},\) and \(t=2.0 \mathrm{s} .\) (c) Sketch the dot's trajectory from \(t=0\) to \(t=2.0 \mathrm{s},\) and show the velocities calculated in part (b).

A "moving sidewalk" in an airport terminal building moves at 1.0 \(\mathrm{m} / \mathrm{s}\) and is 35.0 \(\mathrm{m}\) long. If a woman steps on at one end and walks at 1.5 \(\mathrm{m} / \mathrm{s}\) relative to the moving sidewalk, how much time does she require to reach the opposite end if she walks (a) in the same direction the sidewalk is moving? (b) In the opposite direction?

An airplane pilot sets a compass course due west and maintains an airspeed of 220 \(\mathrm{km} / \mathrm{h}\) . After flying for 0.500 \(\mathrm{h}\) , she finds herself over a town 120 \(\mathrm{km}\) west and 20 \(\mathrm{km}\) south of her starting point. (a) Find the wind velocity (magnitude and direction). (b) If the wind velocity is 40 \(\mathrm{km} / \mathrm{h}\) due south, in what direction should the pilot set her course to travel due west? Use the same airspeed of 220 \(\mathrm{km} / \mathrm{h}\) .

A faulty model rocket moves in the \(x y\) -plane (the positive \(y\) -direction is vertically upward. The rocket's acceleration has components \(a_{x}(t)=\alpha t^{2}\) and \(a_{y}(t)=\beta-\gamma t,\) where \(\alpha=2.50 \mathrm{m} / \mathrm{s}^{4}, \beta=9.00 \mathrm{m} / \mathrm{s}^{2},\) and \(\gamma=1.40 \mathrm{m} / \mathrm{s}^{3} .\) At \(t=0\) the rocket is at the origin and has velocity \(\vec{\boldsymbol{v}}_{0}=v_{0 x} \hat{\boldsymbol{\imath}}+v_{0 y} \hat{\boldsymbol{J}}\) with \(v_{0 x}=1.00 \mathrm{m} / \mathrm{s}\) and \(v_{0 y}=7.00 \mathrm{m} / \mathrm{s} .\) (a) Calculate the velocity and position vectors as functions of time. (b) What is the maximum height reached by the rocket? (c) Sketch the path of the rocket. (d) What is the horizontal displacement of the rocket when it returns to \(y=0 ?\)
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