Animated Solution for Physics - Kinematics: A butterfly is flying with a velocity $4\sqrt{2}$ m/s in North-East direction. Wind is slowly blowing at 1 m/s from North to South. The resultant displacement of the butterfly in 3 s is

Imagine a butterfly fluttering in the air. It is trying its best to fly North-East with a speed of $4\sqrt{2}\text{ m/s}$. But there is a catch—the air itself is moving! This means the butterfly's given velocity is actually its velocity relative to the wind.

To find out where the butterfly actually ends up, we need its absolute velocity with respect to the ground. Let's call this absolute velocity ${\mathbf{v}}_2$. By the fundamental concept of relative velocity, the velocity of the butterfly relative to the wind, ${\mathbf{v}}_{21}$, is equal to ${\mathbf{v}}_2-{\mathbf{v}}_1$, where ${\mathbf{v}}_1$ is the wind's velocity.

Therefore, to find ${\mathbf{v}}_2$, we simply rearrange the equation:

First, let's break down the butterfly's relative velocity into its horizontal ($x$) and vertical ($y$) components. Since it is flying North-East, the angle it makes with the East direction is exactly $45^{\circ }$.

Using basic trigonometry, we can write:

We know that both $\cos 45^{\circ }$ and $\sin 45^{\circ }$ are equal to $\frac{1}{\sqrt{2}}$. When we substitute this value, the $\sqrt{2}$ terms cancel out perfectly! We are left with a neat and simple vector:

Next, let's look at the wind. The problem states the wind is blowing from North to South at $1\text{ m/s}$. South corresponds to the negative $y$-direction. So, the wind's velocity vector, ${\mathbf{v}}_1$, is simply:

Now for the magic! To find the butterfly's actual velocity over the ground, ${\mathbf{v}}_2$, we add the relative velocity and the wind velocity together.

The $\hat{\mathbf{i}}$ component stays the same, but $4\hat{\mathbf{j}}-1\hat{\mathbf{j}}$ gives us $3\hat{\mathbf{j}}$. So, the absolute velocity is:

We have the absolute velocity, but we need the displacement after $3\text{ seconds}$. Displacement is simply velocity multiplied by time (since the velocity is constant).

This gives us a displacement vector of:

Finally, we need the magnitude of this displacement. We use the Pythagorean theorem:

And the square root of $225$ is exactly $15\text{ m}$!

This is a classic example of how a medium affects the motion of an object within it. Whether it is a butterfly in the wind, a swimmer in a river, or an airplane in the jet stream, the absolute motion is always the vector sum of the relative motion and the medium's motion.