Animated Solution for Physics - Kinematics: A particle is moving Eastwards with a velocity of $5$ m/s. In $10$ s, the velocity changes to $5$ m/s Northwards. The average acceleration in this time is

The journey of mastering physics often begins with a profound realization: the world is not just about how much, but also about which way. This problem is a classic, beautiful trap designed to test exactly that realization. It looks innocent, almost too easy, but it hides a fundamental truth about the nature of motion.

Imagine you are driving a car. You are cruising along a straight highway heading East at a steady speed of $5\text{ m/s}$. You feel relaxed; the ride is smooth. Suddenly, the road curves sharply, and within $10\text{ s}$, you find yourself heading North, still maintaining that exact same speed of $5\text{ m/s}$.

A casual observer might look at your speedometer and say, "Hey, your speed didn't change! It was $5\text{ m/s}$ before, and it's $5\text{ m/s}$ now. Therefore, your acceleration must be zero."

But as a student of physics, you know better. You know that to change your direction, you had to turn the steering wheel. The tires had to grip the road, exerting a force to push you Northward while stopping your Eastward momentum. And where there is a net force, there must be an acceleration. This is the essence of vectors.

Let us formalize our intuition. We are dealing with a particle that has an initial state and a final state.

First, we define our coordinate system. Let the East direction be represented by the positive x-axis (unit vector $\hat{i}$) and the North direction by the positive y-axis (unit vector $\hat{j}$).

The initial velocity of the particle is given as $5\text{ m/s}$ towards the East. We can write this mathematically as:

After a time interval of $\Delta t=10\text{ s}$, the particle's velocity changes. It is now moving Northwards with the same speed. We write the final velocity as:

The question asks for the average acceleration during this time interval.

To find the average acceleration, we must reach into our physics arsenal and pull out the fundamental definition. Average acceleration is defined as the total change in velocity divided by the total time taken.

This equation is simple, but the numerator, $\Delta \mathbf{v}$, is where the magic happens. The change in velocity is the final velocity minus the initial velocity:

Here is the critical trap: we cannot simply subtract the magnitudes. $5-5=0$ is the wrong answer because velocity is a vector. We must perform vector subtraction.

Geometrically, subtracting a vector is the same as adding its negative.

What does $-{\mathbf{v}}_i$ look like? If ${\mathbf{v}}_i$ points East, then $-{\mathbf{v}}_i$ must point exactly in the opposite direction, which is West. Its magnitude remains $5\text{ m/s}$.

So, our task boils down to adding a Northward vector (${\mathbf{v}}_f$) and a Westward vector ($-{\mathbf{v}}_i$).

Imagine placing the tail of the Westward vector at the origin. Now, place the tail of the Northward vector at the origin as well. To find their sum, we can use the parallelogram law of vector addition. Since the vectors point North and West, the angle between them is exactly $90^{\circ }$.

The resultant vector, $\Delta \mathbf{v}$, will point diagonally between North and West. Because both vectors have the exact same magnitude ($5\text{ m/s}$), the resultant will perfectly bisect the $90^{\circ }$ angle. This means the direction of the change in velocity is exactly North-West.

Now, let's calculate the magnitude of this change in velocity. Since the vectors form a right-angled triangle, we can use the Pythagorean theorem:

Substituting our values:

So, the change in velocity is a vector of magnitude $5\sqrt{2}\text{ m/s}$ pointing North-West.

We are now in the home stretch. We have the change in velocity, and we have the time interval. We just need to plug these back into our master equation for average acceleration.

Substituting the magnitude and the time $\Delta t=10\text{ s}$:

Let's simplify this fraction. Dividing the numerator and the denominator by $5$, we get:

To make this match the options provided in the question, we can rewrite the denominator $2$ as $\sqrt{2}\times \sqrt{2}$:

Canceling one $\sqrt{2}$ from the top and bottom leaves us with our final, elegant result:

And what about the direction? The acceleration vector always points in the exact same direction as the change in velocity vector ($\Delta \mathbf{v}$). Since we established that $\Delta \mathbf{v}$ points North-West, the average acceleration must also point North-West.

Therefore, the average acceleration is $\frac{1}{\sqrt{2}}{\text{ m/s}}^2$ towards the North-West. This perfectly matches option (c).

This problem is a beautiful reminder that in physics, direction matters just as much as magnitude. A change in direction is a change in velocity, and a change in velocity requires acceleration. Never let a constant speed fool you into thinking nothing is happening!