Problem 14 - Fundamentals of Physics

Question

You are to make four straight-line moves over a flat desert floor, starting at the origin of an \(x y\) coordinate system and ending at the \(x y\) coordinates \((-140 \mathrm{~m}, 30 \mathrm{~m})\). The \(x\) component and \(y\) component of your moves are the following, respectively, in meters: \((20\) and 60\()\), then \(\left(b_{x}\right.\) and \(\left.-70\right)\), then \(\left(-20\right.\) and \(\left.c_{y}\right)\), then \((-60\) and \(-70\) ). What are (a) component \(b_{x}\) and (b) component \(c_{y}\) ? What are (c) the magnitude and (d) the angle (relative to the positive direction of the \(x\) axis) of the overall displacement?

Step-by-Step Solution

Answer

(a) \(b_x = -100\), (b) \(c_y = 110\), (c) Magnitude \(\approx 143.17 \text{ m}\), (d) Angle \(\approx 167.89°\).

Step 1: Set up the total displacement equation

You know your total displacement resulting from four moves ends at the point \((-140, 30)\). You can set up the displacement equation for both the \(x\) and \(y\) components:\[x_{total} = 20 + b_x - 20 - 60 = -140\]\[y_{total} = 60 - 70 + c_y - 70 = 30\]

Step 2: Solve for component b_x

Let's solve the equation we established for the \(x\) components:\[20 + b_x - 20 - 60 = -140\]Combine like terms:\[b_x - 60 = -160\]Now solve for \(b_x\):\[b_x = -160 + 60 = -100\]

Step 3: Solve for component c_y

Next, solve the equation for the \(y\) components:\[60 - 70 + c_y - 70 = 30\]Combine like terms:\[-80 + c_y = 30\]Now solve for \(c_y\):\[c_y = 30 + 80 = 110\]

Step 4: Calculate the magnitude of the displacement

The overall displacement is from the origin \((0,0)\) to \((-140, 30)\), which can be calculated using the Pythagorean theorem:\[d = \sqrt{(-140)^2 + (30)^2}\]\[d = \sqrt{19600 + 900} = \sqrt{20500}\]\[d = approximately \; 143.17\text{ m}\]

Step 5: Calculate the angle of displacement

The angle \(\theta\) relative to the positive \(x\)-axis is determined using the tangent function:\[\theta = \arctan\left(\frac{y_{total}}{x_{total}}\right)\]\[\theta = \arctan\left(\frac{30}{-140}\right)\]\[\theta \approx \arctan\left(-0.2143\right)\]Since the resulting \(x\)-component is negative and the \(y\)-component is positive, \(\theta\) is in the second quadrant. The angle is:\[\theta \approx 180° - 12.11° = 167.89°\]

Key Concepts

Coordinate System

A coordinate system allows us to precisely describe the location of a point in space. The most common system is the Cartesian coordinate system, which uses two perpendicular axes: the x-axis (horizontal) and the y-axis (vertical).
Each point in this system is represented as an ordered pair \(x, y\). The origin, denoted as \(0, 0\), is where these axes intersect. It's the starting point for our displacement in this problem.
When dealing with vector displacement, we begin from one point and move to another, like traveling from the origin to the final point \(-140, 30\).
This kind of problem will involve tracking each move as a change in both x and y directions to find the final position.

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry, particularly useful in calculating the magnitude of vectors.
When applied in a coordinate system, this theorem helps find the length of a vector, or displacement, when you know its horizontal and vertical components.
For a right triangle formed by a vector in an \(xy\)-plane, the relationship between the sides is \(a^2 + b^2 = c^2\), where \(a\) and \(b\) are the legs and \(c\) is the hypotenuse or the vector's magnitude.
In our case, the components \(-140\) and \(+30\) are used as sides of a right triangle to calculate the resultant displacement, \(d\). It's done as \( d = \sqrt{(-140)^2 + (30)^2}\).
This approach allows you to measure the actual distance moved from the starting point to the final location.

Trigonometry

Trigonometry deals with the relationships between angles and sides in triangles, and is essential in vector calculations.
In a coordinate system, trigonometric functions like sine, cosine, and tangent help find angles and lengths when vectors are involved.
For example, when calculating the angle of overall displacement, we use the tangent function:
\[\tan(\theta) = \frac{y_{total}}{x_{total}}\]
This equation utilizes the total magnitude of the components to find the angle \(\theta\). It becomes crucial to understand which quadrant the angle lies to provide the correct direction. In this problem, the resulting angle lies in the second quadrant because the x-component is negative while the y-component is positive.

Angle Calculation

Calculating angles involves determining the direction of a vector relative to a reference axis, usually the positive x-axis.
To find an angle in a vector problem, the tangent function is often used as it relates the opposite and adjacent sides of a right triangle.
The formula \(\theta = \arctan\left(\frac{y_{total}}{x_{total}}\right)\) allows us to find the angle formed by the vector and the x-axis.
In our scenario, the calculation gives \(\theta \approx \arctan(-0.2143)\).
The negative x-component indicates the vector is pointing to the left, placing the angle in the second quadrant. Hence, the angle is calculated as \(\theta \approx 180° - 12.11° = 167.89°\).
This precise calculation helps establish the overall direction of the displacement.