Animated Solution for Physics - Kinematics: A particle moves from the point $(2.0\hat{i}+4.0\hat{j})\text{ m}$ at $t=0$ with an initial velocity $(5.0\hat{i}+4.0\hat{j}){\text{ ms}}^{-1}$. It is acted upon by a constant force which produces a constant acceleration $(4.0\hat{i}+4.0\hat{j}){\text{ ms}}^{-2}$. What is the distance of the particle from the origin at time $2\text{ s}$?

Imagine you are standing on a massive, perfectly flat parking lot. You are not standing at the exact center (the origin); instead, you are a few steps away. Suddenly, you start running, and a steady, unyielding wind begins to push you, constantly accelerating your motion. This physical scenario is exactly what our particle is experiencing in the 2D plane. Let's break down the mathematics of this journey and see how vector kinematics elegantly solves the problem.

Not Starting at Zero

The most common trap in kinematics problems is assuming that everything begins at the origin $(0,0)$. Our problem explicitly states that at $t=0$, the particle is at a specific coordinate:

This is our initial position vector. It acts as the anchor point for all future movement. The particle also has an initial velocity, meaning it's already moving when we start our stopwatch:

The problem mentions that the particle is acted upon by a constant force. According to Newton's Second Law ($\mathbf{F}=m\mathbf{a}$), a constant force guarantees a constant acceleration. This is the golden key that unlocks the standard kinematic equations. We are given:

Because the acceleration is constant, we don't need to perform complex calculus or integrations. We can directly deploy the vector form of the second equation of motion. This equation is incredibly powerful because it handles both the X and Y dimensions simultaneously:

Notice how this equation is structured. It says: "Your final position is your starting position, plus the distance you would cover if you just kept moving at your initial speed, plus the extra distance you cover because you are accelerating."

We need to find the position at $t=2\text{ s}$. Let's substitute our known vectors into the master equation:

Now, we execute the scalar multiplications carefully. For the velocity term, we multiply by $2$. For the acceleration term, $t^2=4$, and multiplying by $\frac{1}{2}$ gives a scalar factor of $2$:

The beauty of vector addition is that the $\hat{i}$ components (horizontal motion) and $\hat{j}$ components (vertical motion) are completely independent. We simply group the like terms together:

This is the final position vector of the particle. It tells us that after 2 seconds, the particle is located at the coordinates $(20,20)$.

The question doesn't just ask for the coordinates; it asks for the distance from the origin. Geometrically, the position vector $\mathbf{r}$ forms the hypotenuse of a right-angled triangle with the X and Y axes. To find its length (magnitude), we use the Pythagorean theorem:

Substituting our final coordinates:

And there we have it! By trusting the vector equations and carefully managing our components, we've tracked the particle's parabolic journey across the plane to find its exact distance from the origin.