Animated Solution for Physics - Kinematics: In $1.0\text{ s}$, a particle goes from point $A$ to point $B$, moving in a semicircle (see figure). The magnitude of the average velocity is

Imagine you are standing at the edge of a perfectly circular track. You start at point $A$, the very top of the track, and you run along the curved boundary until you reach point $B$, exactly on the opposite side. You have just traced out a perfect semicircle.

This is the exact physical reality our particle is experiencing in this problem. It moves along a semicircular path from point $A$ to point $B$. The problem provides us with two crucial pieces of information: the radius of this circular path is $R=1.0\text{ m}$, and the total time taken for this journey is $\Delta t=1.0\text{ s}$.

Now, the question asks us for the magnitude of the average velocity. This is where the trap is set. The examiners are testing your fundamental understanding of kinematics. They want to see if you will blindly calculate the length of the path you ran, or if you will stop and think about what velocity actually means in physics.

Many students, when they see a curved path, immediately start thinking about circumferences and arc lengths. They see the radius, they see the semicircle, and their brain automatically jumps to the formula for the perimeter of a circle. But physics is not just about plugging numbers into the first formula that comes to mind. It is about understanding the physical meaning behind the words.

Displacement: The Core Difference

To solve this problem, we must confront one of the most classic misconceptions in physics: the fundamental difference between distance and displacement. This distinction is the cornerstone of kinematics, and mastering it is essential for any serious physics student.

Distance is a scalar quantity. It represents the actual length of the path traveled by the particle. It is the reading on your car's odometer. If you were to lay a piece of string along the semicircular curve from $A$ to $B$ and then measure that string with a ruler, you would be measuring the distance.

In this specific case, the distance would be half the circumference of the circle, which is mathematically expressed as $\pi R$. If we calculate this using our given radius, we get $3.14\times 1.0=3.14\text{ m}$. Notice how $3.14\text{ m/s}$ is cleverly placed as option (a) in the multiple-choice answers! This is a deliberate trap designed to catch students who calculate average speed instead of average velocity.

Displacement, on the other hand, is a vector quantity. It is a much more rigid and unforgiving concept. Displacement cares absolutely nothing about the scenic route you took. It ignores the curves, the detours, and the actual path length. It only cares about two things: where you started and where you ended up.

Displacement is defined strictly as the shortest straight-line distance from the initial position to the final position, pointing from the start to the end. It is the "as the crow flies" distance.

Let's look closely at our geometric setup. The particle starts its journey at point $A$ and ends its journey at point $B$. What is the absolute shortest straight-line path between these two specific points?

Because $A$ and $B$ are located on exactly opposite ends of a perfect semicircle, the straight line connecting them must pass directly through the center of the circle. By the fundamental definitions of geometry, a line segment that connects two points on a circle and passes straight through the center is the diameter of the circle.

Therefore, the magnitude of the displacement vector is exactly equal to the diameter of the circle. Since the diameter is always twice the radius, we can write this relationship mathematically as:

We know from the problem statement that the radius $R$ is exactly $1.0\text{ m}$. Let's carefully substitute this value into our equation to find the exact numerical value of the displacement:

So, even though the particle traveled a longer distance of $3.14\text{ m}$ along the curved arc, its actual change in position—its true displacement—is only $2.0\text{ m}$ straight down. This is a profound realization. The particle did a lot of extra moving that didn't contribute to its final net change in position!

Average Velocity

Now that we have successfully calculated our displacement, we are ready to deploy our master equation. Average velocity is defined fundamentally as the total displacement divided by the total time taken for the journey.

This equation is the absolute heart of the problem. It elegantly connects the spatial geometry we just analyzed—the displacement—with the temporal information given in the problem—the time taken.

It is crucial to note that because displacement is a vector, average velocity is also a vector. The direction of the average velocity is always exactly the same as the direction of the displacement. In this case, it points straight from $A$ to $B$. However, the question only asks for the magnitude of the average velocity, so we only need to concern ourselves with the numerical values.

We now have all the pieces of the puzzle assembled. We know the magnitude of the displacement is exactly $2.0\text{ m}$, and the problem explicitly states that the time taken for the entire semicircular journey is $1.0\text{ s}$.

Let's substitute these known values into our master equation for average velocity:

Performing this simple division yields our final, elegant result:

The magnitude of the average velocity is exactly $2.0\text{ m/s}$. Looking at our multiple-choice options, this perfectly matches option (b).

This problem serves as a beautiful and essential reminder that in physics, precise definitions matter immensely. By staying true to the strict vector definition of displacement and resisting the urge to calculate the arc length, we easily avoided the examiner's trap and arrived at the correct, mathematically sound solution. Always trust the fundamental definitions!