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 average velocity has \(x\)-component -3.8 m/s and y-component 4.9 m/s. At time \(t_2\) = 12.0 s, (a) what are the \(x\)- and \(y\)-coordinates of the rhino? (b) How far is the rhino from the origin?
Short Answer
Step by step solution
Understanding Average Velocity
Finding Displacement in x-direction
Finding Displacement in y-direction
Calculating Final Coordinates
Calculating Distance from Origin
Simplifying the Distance Calculation
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Average Velocity
When we talk about average velocity in multiple dimensions, like in this exercise, we break it down into components. Here, the velocity along the x-axis is \(v_x = -3.8 \, \text{m/s}\) and along the y-axis is \(v_y = 4.9 \, \text{m/s}\). These values indicate that over 12 seconds, the average motion in the x-direction is negative, which means it's moving to the left, while in the y-direction, it's positive, indicating upwards movement.
- Average velocity involves both magnitude and direction.
- It can be visualized as the slope of a line on a graph of position versus time.
- Use the formula \( \text{Average Velocity} = \frac{\Delta \text{Position}}{\Delta \text{Time}} \).
Displacement Calculation
In our rhino scenario, we determine displacement separately for the x and y directions.
- For the x-direction: \( \Delta x = v_x \times \Delta t = -3.8 \, \text{m/s} \times 12.0 \, \text{s} = -45.6 \, \text{m} \).
- For the y-direction: \( \Delta y = v_y \times \Delta t = 4.9 \, \text{m/s} \times 12.0 \, \text{s} = 58.8 \, \text{m} \).
Pythagorean Theorem
In a right triangle, the theorem states that 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 formula terms: \[ c^2 = a^2 + b^2 \].
In our scenario, we use this theorem to find the straight-line distance from the rhino's starting point at the origin to its new location. We plug in the displacements:
- Calculate: \( 2079.36 = (-45.6)^2 \).
- Calculate: \( 3457.44 = (58.8)^2 \).
- Sum these: \( 5536.80 \).
- Find the square root: \( \sqrt{5536.80} \approx 74.37 \, \text{m} \).
Coordinate Geometry
In this exercise, the rhino's movement over time is given in terms of coordinates. The initial location is the origin, \((0,0)\), and you calculate its final position based on its displacement along the x and y axes:
- Final x-coordinate: \(-45.6 \).
- Final y-coordinate: \(58.8 \).
- The coordinates of the rhino after 12 seconds, therefore, are \((-45.6, 58.8)\).
Coordinate geometry helps us visualize these spatial changes, making it easier to interpret movement and positions in a visual framework.