Problem 12 - Fundamentals of Physics

Question

At one instant a bicyclist is \(40.0 \mathrm{~m}\) due east of a park's flagpole, going due south with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\). Then \(30.0 \mathrm{~s}\) later, the cyclist is \(40.0 \mathrm{~m}\) due north of the flagpole, going due east with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\). For the cyclist in this \(30.0\) s interval, what are the (a) magnitude and (b) direction of the displacement, the (c) magnitude and (d) direction of the average velocity, and the (e) magnitude and (f) direction of the average acceleration?

Step-by-Step Solution

Answer

Magnitude of displacement: 56.57 m; Direction of displacement: 135°; Average velocity: 1.88 m/s at 135°; Average acceleration: 0.47 m/s² at 45°.

Step 1: Determine Initial and Final Positions

Initially, the bicyclist is located 40 m east of the flagpole and travels due south. After 30 s, the bicyclist is 40 m north of the flagpole. The displacement is from the initial position to the final position, forming a straight line from 40 m east to 40 m north.

Step 2: Calculate Displacement

To find the displacement, consider the change along the horizontal (east-west) and vertical (north-south) directions. Initially at (40, 0), and finally at (0, 40), the displacement is given by the vector difference \(\Delta \vec{r} = (0 - 40, 40 - 0) = (-40, 40)\). The magnitude of the displacement is \(\sqrt{(-40)^2 + 40^2} = \sqrt{3200} = 56.57\,\mathrm{m}\).

Step 3: Determine Direction of Displacement

The direction of the displacement vector \((-40, 40)\) can be calculated using \(\tan^{-1}\left(\frac{40}{-40}\right) = \tan^{-1}(-1)\). This yields an angle of -45 degrees, which points from the final over the initial position, corresponding to 135 degrees relative to east (counterclockwise).

Step 4: Calculate Average Velocity

The average velocity \(\vec{v}_{\text{avg}}\) is the ratio of the displacement to the time interval. Therefore, \(\vec{v}_{\text{avg}} = \frac{(-40, 40)}{30}\). This results in \(\vec{v}_{\text{avg}} = (-1.33, 1.33)\,\mathrm{m/s}\). The magnitude is \(\sqrt{(-1.33)^2 + 1.33^2} = 1.88\,\mathrm{m/s}\).

Step 5: Determine Direction of Average Velocity

The direction for the average velocity vector \((-1.33, 1.33)\) is the same as that for displacement. Hence, the direction angle is 135 degrees relative to east.

Step 6: Calculate Change in Velocity

The initial velocity is due south at 10 m/s, which translates to \((0, -10)\,\mathrm{m/s}\). The final velocity is due east at 10 m/s, which is \((10, 0)\,\mathrm{m/s}\). The change in velocity \(\Delta \vec{v}\) is \((10, 0) - (0, -10) = (10, 10)\).

Step 7: Calculate Average Acceleration

The average acceleration \(\vec{a}_{\text{avg}}\) is the change in velocity divided by the time interval. Hence, \(\vec{a}_{\text{avg}} = \frac{(10, 10)}{30} = (0.33, 0.33)\,\mathrm{m/s^2}\). The magnitude is \(\sqrt{0.33^2 + 0.33^2} = 0.47\,\mathrm{m/s^2}\).

Step 8: Determine Direction of Average Acceleration

The direction of the average acceleration \((0.33, 0.33)\) follows the vector's angle, calculated as \(\tan^{-1}\left(\frac{0.33}{0.33}\right) = \tan^{-1}(1) = 45\) degrees relative to east.

Key Concepts

Displacement

Displacement is a fundamental concept in kinematics. It refers to the change in position of an object and is represented as a vector quantity. This means it has both magnitude and direction. Unlike distance, which only considers how much ground an object has covered, displacement considers the shortest path between the initial and final positions.
In the given problem, the bicyclist's journey is described step-by-step in terms of displacement. Initially, the bicyclist is 40 meters east of a flagpole and ends up 40 meters north after 30 seconds. To find the displacement, we must determine the straight-line distance and direction from their starting to ending points.
The displacement vector is calculated by subtracting the initial position vector from the final position vector. For this scenario, the displacement is \((-40, 40)\) meters which forms a diagonal vector on the coordinate plane. The magnitude of this vector, representing the actual distance of displacement, is calculated as \(\sqrt{(-40)^2 + 40^2} = 56.57 \,\text{m}\). The direction is determined using trigonometric functions, giving an angle of 135 degrees relative to east.

Average Velocity

Average velocity is another crucial aspect of kinematics. It refers to the total displacement divided by the total time taken. Since it is a vector, it considers both magnitude and direction. This is different from average speed, which is purely about distance and not direction.
To find the average velocity in our bicyclist situation, we use the displacement calculated earlier and divide it by the time period of 30 seconds. The average velocity vector is \((-1.33, 1.33) \,\text{m/s}\). This shows that the cyclist's overall direction encapsulated both eastward and northward movements during the interval.
The magnitude of the average velocity gives us the speed of travel in a straight line from the starting point to the endpoint. For this problem, that is approximately \(1.88 \,\text{m/s}\). Finally, since the direction of average velocity derives from the displacement vector, its direction is also 135 degrees relative to east, showing a balance of northward and eastward components in the travel path.

Average Acceleration

Average acceleration is defined as the change in velocity over the change in time. It, too, is a vector quantity. It captures how quickly the velocity of an object is changing over a period of time. In simple terms, it is a measure of how speed or direction or both change with time.
For our bicyclist, the initial velocity is southward at 10 m/s, denoted as the vector \(0, -10) \,\text{m/s}\), and the final velocity is eastward at 10 m/s, written as \(10, 0) \,\text{m/s}\). Hence, the change in velocity vector is \(10, 10\), which shows the total shift in speed and direction.
To find the average acceleration, divide this change in velocity by the 30 seconds time interval. The resulting average acceleration vector is \(0.33, 0.33) \,\text{m/s}^2\), with a magnitude of approximately \(0.47 \,\text{m/s}^2\). The direction is calculated to be 45 degrees relative to east, indicating the combined eastward acceleration and northward components.

Vector Calculations

Understanding vector calculations is essential when dealing with kinematics. Vectors are mathematical entities that have both magnitude and direction, making them perfect for representing quantities like displacement, velocity, and acceleration.
Calculating vectors involves considering components in multiple directions, typically on an x-y plane. Each vector can be broken down into its horizontal (x) and vertical (y) components. To add or subtract vectors, you work separately on each component.
In our exercise, knowing how to handle vectors helped solve the problem. Displacement, average velocity, and acceleration were expressed as vectors in terms of their components. Their magnitudes were found using the Pythagorean theorem: \(\sqrt{x^2 + y^2}\). Directions were deduced through trigonometry by calculating the angle with the horizontal axis using the \(\tan^{-1}\) function.
By utilizing these operations, we can thoroughly understand the motion described in the problem, making calculated, logical conclusions about movements in real-world contexts.