Answer :
Sure, let's go through a detailed, step-by-step solution to show that the equation [tex]\(\Delta x = v_i t + \frac{1}{2} a t^2\)[/tex] represents the position of a particle moving in a horizontal direction with constant acceleration.
1. Identify the variables:
- [tex]\(\Delta x\)[/tex] is the displacement of the particle from its initial position.
- [tex]\(v_i\)[/tex] is the initial velocity of the particle.
- [tex]\(t\)[/tex] is the time elapsed.
- [tex]\(a\)[/tex] is the constant acceleration of the particle.
2. Understand the initial conditions:
- Assume the initial position of the particle at [tex]\(t = 0\)[/tex] is [tex]\(x_0\)[/tex].
- For simplicity, we can take [tex]\(x_0 = 0\)[/tex]. This means we consider the initial position to be the origin.
3. Use the kinematic equation:
In physics, the equation of motion for a particle moving with constant acceleration is given by:
[tex]\[ x = x_0 + v_i t + \frac{1}{2} a t^2 \][/tex]
Since we assumed [tex]\(x_0 = 0\)[/tex], the equation simplifies to:
[tex]\[ x = v_i t + \frac{1}{2} a t^2 \][/tex]
4. Define the displacement [tex]\(\Delta x\)[/tex]:
The displacement [tex]\(\Delta x\)[/tex] is the change in position of the particle. It is given by:
[tex]\[ \Delta x = x - x_0 \][/tex]
Since [tex]\(x_0 = 0\)[/tex], this further simplifies to:
[tex]\[ \Delta x = x \][/tex]
5. Substitute [tex]\(x\)[/tex] from the equation of motion:
From the simplified equation of motion [tex]\( x = v_i t + \frac{1}{2} a t^2 \)[/tex], we have:
[tex]\[ \Delta x = x = v_i t + \frac{1}{2} a t^2 \][/tex]
Thus, we have shown that the displacement [tex]\(\Delta x\)[/tex] of a particle moving in a horizontal direction with constant acceleration is given by:
[tex]\[ \Delta x = v_i t + \frac{1}{2} a t^2 \][/tex]
This represents the position of the particle at any time [tex]\(t\)[/tex], considering that the initial position was zero and the particle is under constant acceleration [tex]\(a\)[/tex].
1. Identify the variables:
- [tex]\(\Delta x\)[/tex] is the displacement of the particle from its initial position.
- [tex]\(v_i\)[/tex] is the initial velocity of the particle.
- [tex]\(t\)[/tex] is the time elapsed.
- [tex]\(a\)[/tex] is the constant acceleration of the particle.
2. Understand the initial conditions:
- Assume the initial position of the particle at [tex]\(t = 0\)[/tex] is [tex]\(x_0\)[/tex].
- For simplicity, we can take [tex]\(x_0 = 0\)[/tex]. This means we consider the initial position to be the origin.
3. Use the kinematic equation:
In physics, the equation of motion for a particle moving with constant acceleration is given by:
[tex]\[ x = x_0 + v_i t + \frac{1}{2} a t^2 \][/tex]
Since we assumed [tex]\(x_0 = 0\)[/tex], the equation simplifies to:
[tex]\[ x = v_i t + \frac{1}{2} a t^2 \][/tex]
4. Define the displacement [tex]\(\Delta x\)[/tex]:
The displacement [tex]\(\Delta x\)[/tex] is the change in position of the particle. It is given by:
[tex]\[ \Delta x = x - x_0 \][/tex]
Since [tex]\(x_0 = 0\)[/tex], this further simplifies to:
[tex]\[ \Delta x = x \][/tex]
5. Substitute [tex]\(x\)[/tex] from the equation of motion:
From the simplified equation of motion [tex]\( x = v_i t + \frac{1}{2} a t^2 \)[/tex], we have:
[tex]\[ \Delta x = x = v_i t + \frac{1}{2} a t^2 \][/tex]
Thus, we have shown that the displacement [tex]\(\Delta x\)[/tex] of a particle moving in a horizontal direction with constant acceleration is given by:
[tex]\[ \Delta x = v_i t + \frac{1}{2} a t^2 \][/tex]
This represents the position of the particle at any time [tex]\(t\)[/tex], considering that the initial position was zero and the particle is under constant acceleration [tex]\(a\)[/tex].
