Animated Solution for Physics - Kinematics: A particle is moving Eastwards with a velocity of $5{\text{ ms}}^{-1}$. In $10\text{ s}$, the velocity changes to $5{\text{ ms}}^{-1}$ Northwards. The average acceleration in this time is

Imagine you are driving a car on a perfectly flat, open ground. You are cruising exactly Eastwards at a steady speed of $5{\text{ ms}}^{-1}$. Suddenly, you decide to take a turn, and $10\text{ s}$ later, you find yourself moving exactly Northwards, still at the same steady speed of $5{\text{ ms}}^{-1}$.

A common trap here is to think: "My speed didn't change, so my acceleration must be zero!"

But remember, velocity is a vector. It cares just as much about where you are going as it does about how fast you are going. A change in direction is a change in velocity, and any change in velocity requires an acceleration.

Let's translate our physical reality into the language of mathematics using unit vectors. We will assign $\hat{i}$ to the East direction and $\hat{j}$ to the North direction.

Our initial velocity is:

Our final velocity is:

To find the average acceleration, we need the master equation of kinematics:

The numerator, $\Delta \mathbf{v}$, is the vector difference ${\mathbf{v}}_2-{\mathbf{v}}_1$. Geometrically, subtracting a vector is identical to adding its opposite. So, we are effectively calculating ${\mathbf{v}}_2+(-{\mathbf{v}}_1)$.

If ${\mathbf{v}}_1$ points East, then $-{\mathbf{v}}_1$ points West.

This new vector, $\Delta \mathbf{v}$, has a negative x-component and a positive y-component. If you draw it out, it points diagonally into the second quadrant—exactly North-West.

Now, let's find the magnitude of this change in velocity using the Pythagorean theorem:

Finally, we divide this magnitude by the time interval $\Delta t=10\text{ s}$ to find the magnitude of the average acceleration:

Since the acceleration vector is simply the $\Delta \mathbf{v}$ vector scaled down by time, it shares the exact same direction. Therefore, the average acceleration is $\frac{1}{\sqrt{2}}{\text{ ms}}^{-2}$ towards North-West.