Problem 2 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

The rhino's coordinates are (-45.6, 58.8) and it is approximately 74.37 m from the origin.

Step 1: Understanding Average Velocity

Average velocity is the change in position divided by the change in time. Here, the average velocity components are given as \( v_x = -3.8 \text{ m/s} \) and \( v_y = 4.9 \text{ m/s} \) over the time interval \( t_1 = 0 \) to \( t_2 = 12.0 \text{ s} \).

Step 2: Finding Displacement in x-direction

The displacement in the x-direction is calculated using the formula \( \Delta x = v_x \cdot \Delta t \). Here, \( v_x = -3.8 \text{ m/s} \) and \( \Delta t = 12.0 \text{ s} \). Therefore, \( \Delta x = (-3.8 \text{ m/s}) \cdot 12.0 \text{ s} = -45.6 \text{ m} \).

Step 3: Finding Displacement in y-direction

The displacement in the y-direction is calculated using the formula \( \Delta y = v_y \cdot \Delta t \). Here, \( v_y = 4.9 \text{ m/s} \) and \( \Delta t = 12.0 \text{ s} \). Therefore, \( \Delta y = (4.9 \text{ m/s}) \cdot 12.0 \text{ s} = 58.8 \text{ m} \).

Step 4: Calculating Final Coordinates

The rhino starts from the origin, so the final coordinates are given by \( (x, y) = (\Delta x, \Delta y) = (-45.6, 58.8) \).

Step 5: Calculating Distance from Origin

The distance from the origin is found using the Pythagorean theorem \( d = \sqrt{(\Delta x)^2 + (\Delta y)^2} \). Plugging in values, we have \( d = \sqrt{(-45.6)^2 + (58.8)^2} \).

Step 6: Simplifying the Distance Calculation

Calculate \( (-45.6)^2 = 2079.36 \) and \( (58.8)^2 = 3457.44 \). Add these to get \( 5536.80 \), and find the square root to get \( d \approx 74.37 \text{ m} \).

Key Concepts

Average Velocity

Average velocity helps us understand how quickly an object is moving and in which direction over a period of time. It is different from instantaneous velocity, which is the speed and direction at a specific moment. Average velocity takes into account the overall change over a stretch of time.

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

Displacement is a vector quantity that represents the change in position of an object. It's different from distance, which is a scalar quantity indicating "how much ground" an object has covered. Displacement looks at the shortest path from the initial to the final position.

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: \( \Delta y = v_y \times \Delta t = 4.9 \, \text{m/s} \times 12.0 \, \text{s} = 58.8 \, \text{m} \).

These calculations tell us the rhino's new position after it has moved for 12 seconds.

Pythagorean Theorem

The Pythagorean Theorem is a vital mathematical principle for calculating distances. It's especially useful in coordinate geometry for determining how far something is from the origin or between points.

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: \( \sqrt{5536.80} \approx 74.37 \, \text{m} \).

Thus, the rhino's straight-line distance from the origin is approximately 74.37 meters.

Coordinate Geometry

Coordinate geometry, or analytic geometry, combines algebra and geometry to describe geometrical figures with equations. It's a powerful tool for solving real-world problems involving space and shape, particularly with position and movement in 2D and 3D systems.

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.