A Particle Starts From Rest At The Point 2 0

Aparticle starts from rest at the point (2, 0) – Understanding Motion from a Given Initial Position

When a physics problem states that a particle starts from rest at the point (2, 0), it is giving you two crucial pieces of information: the particle’s initial location in a coordinate system and the fact that its initial velocity is zero. This simple statement opens the door to a wide range of kinematic and dynamic analyses, from straight‑line motion with constant acceleration to curved trajectories under varying forces. In this article we will unpack what the phrase means, explore the mathematical tools needed to describe the particle’s subsequent motion, work through several representative examples, and highlight common pitfalls to avoid. By the end, you should feel confident tackling any problem that begins with “a particle starts from rest at the point (2, 0).”

1. Why the Initial Conditions Matter

In classical mechanics, the future state of a particle is determined by its initial position and initial velocity (and, if forces depend on time, the initial acceleration as well). The phrase “starts from rest” tells us that the initial velocity vector (\vec{v}_0) is zero:

[ \vec{v}0 = \langle 0,,0\rangle \quad\text{(in 2‑D)}\qquad\text{or}\qquad v{0x}=v_{0y}=0 . ]

The coordinates ((2,0)) fix the origin of the particle’s trajectory. If we place the coordinate axes in the usual way—(x) horizontal, (y) vertical—the particle begins two units to the right of the origin on the (x)-axis and exactly on the axis itself. Knowing both pieces lets us integrate acceleration to obtain velocity and then position, or, conversely, differentiate a known position function to verify that it satisfies the given initial conditions.

2. Kinematic Equations for Constant Acceleration

The simplest scenario is uniform (constant) acceleration. When (\vec{a}) does not change with time, the motion follows the well‑known kinematic formulas:

[ \begin{aligned} \vec{v}(t) &= \vec{v}_0 + \vec{a},t,\[4pt] \vec{r}(t) &= \vec{r}_0 + \vec{v}_0,t + \tfrac{1}{2}\vec{a},t^{2}. \end{aligned} ]

Because the particle starts from rest, (\vec{v}_0 = \vec{0}), and the equations reduce to:

[ \boxed{\vec{v}(t) = \vec{a},t}\qquad\text{and}\qquad\boxed{\vec{r}(t) = \vec{r}_0 + \tfrac{1}{2}\vec{a},t^{2}} . ]

Here (\vec{r}_0 = \langle 2,,0\rangle). If we know the acceleration vector (perhaps (\vec{a} = \langle a_x, a_y\rangle)), we can instantly write the position as a function of time:

[ \begin{aligned} x(t) &= 2 + \tfrac{1}{2}a_x t^{2},\ y(t) &= 0 + \tfrac{1}{2}a_y t^{2}. \end{aligned} ]

These expressions are linear in the squared time variable, which makes it easy to predict where the particle will be after any given interval.

3. Example 1: Motion Along the x‑Axis with Constant Acceleration

Suppose the only acceleration is a constant (a_x = 4\ \text{m/s}^2) directed to the right, and there is no acceleration in the (y) direction ((a_y = 0)). Plugging into the formulas:

[ \begin{aligned} x(t) &= 2 + \tfrac{1}{2}(4)t^{2} = 2 + 2t^{2},\ y(t) &= 0. \end{aligned} ]

Interpretation: The particle stays on the (x)-axis (since (y(t)=0) for all (t)) and slides farther to the right as time progresses. At (t=1\ \text{s}), (x = 2 + 2(1)^2 = 4\ \text{m}); at (t=2\ \text{s}), (x = 2 + 2(4) = 10\ \text{m}). The velocity grows linearly: (v_x(t) = a_x t = 4t\ \text{m/s}).

4. Example 2: Projectile Motion Launched from (2, 0)

A classic case is a particle thrown from ((2,0)) with an initial speed (v_0) at an angle (\theta) above the horizontal. Even though the problem says “starts from rest,” we can reinterpret it as the particle being released from rest and then acted upon by gravity alone. In that case the initial velocity is still zero, but gravity provides a constant downward acceleration (\vec{a} = \langle 0,,-g\rangle) (with (g\approx 9.81\ \text{m/s}^2)). The motion equations become:

[ \begin{aligned} x(t) &= 2 + \tfrac{1}{2}(0)t^{2} = 2,\ y(t) &= 0 + \tfrac{1}{2}(-g)t^{2} = -\tfrac{1}{2}gt^{2}. \end{aligned} ]

Thus the particle falls straight down from the point ((2,0)). After (t=1\ \text{s}), (y = -4.905\ \text{m}); after (t=2\ \text{s}), (y = -19.62\ \text{m}). The trajectory is a vertical line, which is a useful sanity check: if you ever see a non‑zero (x) displacement in a pure‑gravity problem, you must have missed an initial horizontal velocity component.

5. When Acceleration Is Not Constant

Many real‑world situations involve forces that vary with position or time (e.g., a spring force (\vec{F} = -k\vec{r}) or a drag force proportional to velocity). In those cases we must turn to differential equations:

[ m\vec{a}(t) = m\frac{d^{2}\vec{r}}{dt^{2}} = \vec{F}\big(\vec{r},t\big). ]

With the initial conditions (\vec{r}(0)=\langle 2,0\rangle) and (\vec{v}(0)=\vec{0}), we solve the second‑order ODE. Two common approaches are:

1. Analytical solution – possible for simple forces like Hooke’s law ((F=-kx)), leading to sinusoidal motion.
2. Numerical integration – using methods such as Euler’s or Runge‑Kutta when an analytic form is unavailable.

5.1 Harmonic Oscillator Example

Assume the particle is attached to a spring anchored at the origin, obeying Hooke’s law (\vec{F} = -k\vec{r}). The equation of motion for each component is