Answer :
Sure, let's demonstrate how [tex]\(\Delta x = v_i t + \frac{1}{2} a t^2\)[/tex] is derived and show that it represents the position of a particle moving horizontally with initial velocity [tex]\(v_i\)[/tex] and constant acceleration [tex]\(a\)[/tex].
1. Understanding the Components:
- [tex]\(v_i\)[/tex] is the initial velocity of the particle.
- [tex]\(a\)[/tex] is the constant acceleration.
- [tex]\(t\)[/tex] is the time elapsed.
- [tex]\(\Delta x\)[/tex] is the change in position, i.e., the displacement of the particle after time [tex]\(t\)[/tex].
2. Basic Kinematic Equation:
The basic kinematic equation for uniformly accelerated motion is:
[tex]\[ \Delta x = v_i t + \frac{1}{2} a t^2 \][/tex]
3. Derivation:
To derive this equation, we need two fundamental equations of motion:
a. Velocity as a function of time:
[tex]\[ v = v_i + a t \][/tex]
This equation states that the final velocity [tex]\(v\)[/tex] is the initial velocity [tex]\(v_i\)[/tex] plus the additional velocity gained due to constant acceleration [tex]\(a\)[/tex] over time [tex]\(t\)[/tex].
b. Position as a function of time:
To find the displacement [tex]\(\Delta x\)[/tex], we recognize that the displacement is the area under the velocity-time graph. When the acceleration is constant, this graph is a straight line, and the area under it (displacement) can be calculated as the area of a trapezoid or by integrating the velocity over time.
4. Calculating Displacement:
The displacement [tex]\(\Delta x\)[/tex] can be computed by integrating the velocity equation:
[tex]\[ \Delta x = \int_0^t v \, dt \][/tex]
Substituting [tex]\(v = v_i + a t\)[/tex] into the integral:
[tex]\[ \Delta x = \int_0^t (v_i + a t) \, dt \][/tex]
Breaking it down:
[tex]\[ \Delta x = \int_0^t v_i \, dt + \int_0^t a t \, dt \][/tex]
Solving each integral separately:
[tex]\[ \int_0^t v_i \, dt = v_i \int_0^t 1 \, dt = v_i [t]_0^t = v_i t \][/tex]
[tex]\[ \int_0^t a t \, dt = a \int_0^t t \, dt = a \left[ \frac{t^2}{2} \right]_0^t = a \frac{t^2}{2} = \frac{1}{2} a t^2 \][/tex]
Adding these results together gives:
[tex]\[ \Delta x = v_i t + \frac{1}{2} a t^2 \][/tex]
Therefore, we have shown that the displacement [tex]\(\Delta x\)[/tex] of a particle moving with initial velocity [tex]\(v_i\)[/tex] and constant acceleration [tex]\(a\)[/tex] over time [tex]\(t\)[/tex] is given by the equation:
[tex]\[ \Delta x = v_i t + \frac{1}{2} a t^2 \][/tex]
Using this equation, you can calculate the position of the particle given the initial velocity, constant acceleration, and time elapsed. For example, if [tex]\(v_i = 10 \, \text{m/s}\)[/tex], [tex]\(a = 2 \, \text{m/s}^2\)[/tex], and [tex]\(t = 5 \, \text{s}\)[/tex], the displacement would be [tex]\(75.0 \, \text{m}\)[/tex].
1. Understanding the Components:
- [tex]\(v_i\)[/tex] is the initial velocity of the particle.
- [tex]\(a\)[/tex] is the constant acceleration.
- [tex]\(t\)[/tex] is the time elapsed.
- [tex]\(\Delta x\)[/tex] is the change in position, i.e., the displacement of the particle after time [tex]\(t\)[/tex].
2. Basic Kinematic Equation:
The basic kinematic equation for uniformly accelerated motion is:
[tex]\[ \Delta x = v_i t + \frac{1}{2} a t^2 \][/tex]
3. Derivation:
To derive this equation, we need two fundamental equations of motion:
a. Velocity as a function of time:
[tex]\[ v = v_i + a t \][/tex]
This equation states that the final velocity [tex]\(v\)[/tex] is the initial velocity [tex]\(v_i\)[/tex] plus the additional velocity gained due to constant acceleration [tex]\(a\)[/tex] over time [tex]\(t\)[/tex].
b. Position as a function of time:
To find the displacement [tex]\(\Delta x\)[/tex], we recognize that the displacement is the area under the velocity-time graph. When the acceleration is constant, this graph is a straight line, and the area under it (displacement) can be calculated as the area of a trapezoid or by integrating the velocity over time.
4. Calculating Displacement:
The displacement [tex]\(\Delta x\)[/tex] can be computed by integrating the velocity equation:
[tex]\[ \Delta x = \int_0^t v \, dt \][/tex]
Substituting [tex]\(v = v_i + a t\)[/tex] into the integral:
[tex]\[ \Delta x = \int_0^t (v_i + a t) \, dt \][/tex]
Breaking it down:
[tex]\[ \Delta x = \int_0^t v_i \, dt + \int_0^t a t \, dt \][/tex]
Solving each integral separately:
[tex]\[ \int_0^t v_i \, dt = v_i \int_0^t 1 \, dt = v_i [t]_0^t = v_i t \][/tex]
[tex]\[ \int_0^t a t \, dt = a \int_0^t t \, dt = a \left[ \frac{t^2}{2} \right]_0^t = a \frac{t^2}{2} = \frac{1}{2} a t^2 \][/tex]
Adding these results together gives:
[tex]\[ \Delta x = v_i t + \frac{1}{2} a t^2 \][/tex]
Therefore, we have shown that the displacement [tex]\(\Delta x\)[/tex] of a particle moving with initial velocity [tex]\(v_i\)[/tex] and constant acceleration [tex]\(a\)[/tex] over time [tex]\(t\)[/tex] is given by the equation:
[tex]\[ \Delta x = v_i t + \frac{1}{2} a t^2 \][/tex]
Using this equation, you can calculate the position of the particle given the initial velocity, constant acceleration, and time elapsed. For example, if [tex]\(v_i = 10 \, \text{m/s}\)[/tex], [tex]\(a = 2 \, \text{m/s}^2\)[/tex], and [tex]\(t = 5 \, \text{s}\)[/tex], the displacement would be [tex]\(75.0 \, \text{m}\)[/tex].
