Animated Solution for Physics - Kinematics: A mosquito is moving with a velocity $\mathbf{v}=(0.5t^2\hat{\mathbf{i}}+3t\hat{\mathbf{j}}+9\hat{\mathbf{k}})\text{ m/s}$ and accelerating in uniform conditions. What will be the direction of mosquito after 2s?

Imagine a mosquito buzzing around a room. Unlike a car driving on a straight road, this mosquito has the freedom to move in three dimensions: left/right (X-axis), forward/backward (Y-axis), and up/down (Z-axis). The problem gives us the velocity of this mosquito as a time-dependent vector:

This equation is a mathematical map of the mosquito's speed and direction at any given second. Notice how the $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ components depend on time $t$, meaning the mosquito is accelerating in the XY plane, while its vertical speed (the $\hat{\mathbf{k}}$ component) remains a constant $9\text{ m/s}$.

The Instantaneous Velocity

We are asked to find the direction of the mosquito exactly $2\text{ seconds}$ after it starts moving. To do this, we must freeze time. We substitute $t=2$ into our velocity equation to find the instantaneous velocity vector at that exact moment.

Now we have a static 3D vector. This vector points exactly in the direction the mosquito is flying at the $2$-second mark.

Finding Magnitude

Before we can find the angles this vector makes with the axes, we need to know its total length, or magnitude. In 2D, we use the Pythagorean theorem ($a^2+b^2=c^2$). In 3D, we simply extend this theorem to include the third dimension:

Substituting our components:

So, at $t=2\text{ s}$, the mosquito is flying at a speed of $11\text{ m/s}$.

Navigating 3D Space

To describe the direction of a 3D vector, physicists and mathematicians use Direction Cosines. These are the cosines of the angles the vector makes with the X, Y, and Z axes (usually denoted as $\alpha$, $\beta$, and $\gamma$).

The formula for a direction cosine is beautifully simple: it is the component along that axis divided by the total magnitude of the vector.

Let's find the angle with the X-axis ($\alpha$):

If we look at the options provided in the question, they are all given in terms of ${\tan }^{-1}$. We need to shapeshift our ${\cos }^{-1}$ into a ${\tan }^{-1}$.

Imagine a right-angled triangle where the angle is $\alpha$. Since $\cos \alpha =\frac{\text{Base}}{\text{Hypotenuse}}=\frac{2}{11}$, we can find the Perpendicular using Pythagoras' theorem:

Now, we can write $\tan \alpha =\frac{\text{Perpendicular}}{\text{Base}}=\frac{\sqrt{117}}{2}$. Therefore:

We can repeat this exact same process for the Y-axis angle ($\beta$):

Let's compare our rigorously calculated angles with the options provided: (a) ${\tan }^{-1}\left(\frac{2}{3}\right)$ from X-axis (b) ${\tan }^{-1}\left(\frac{2}{3}\right)$ from Y-axis (c) ${\tan }^{-1}\left(\frac{5}{2}\right)$ from Y-axis (d) ${\tan }^{-1}\left(\frac{5}{2}\right)$ from X-axis

This is a classic scenario in competitive exams where a question might have typographical errors in its options. The official answer key often marks these as 'Bonus' or 'Dropped' questions. However, the true victory here isn't matching an option; it's mastering the flawless mathematical logic required to navigate 3D vectors. You now know exactly how to find the direction of any object moving in three-dimensional space!