Q.Obtain equations of motion for constant acceleration using method of calculus.

The key idea is that instantaneous acceleration is the rate of change of instantaneous velocity, and instantaneous velocity is the rate of change of instantaneous displacement. For constant acceleration $a$, we integrate these definitions.

1. Velocity-Time Relation: Acceleration is defined as $a = \frac{dv}{dt}$. Integrating both sides with initial velocity $u$ at $t=0$ and final velocity $v$ at time $t$:

2. Displacement-Time Relation: Velocity is defined as $v = \frac{ds}{dt}$. Substituting $v = u + at$:

Integrating both sides with initial displacement $0$ at $t=0$ and final displacement $s$ at time $t$:

3. Velocity-Displacement Relation: …

When an object moves, its position, velocity, and acceleration are related. Calculus provides the precise mathematical tools to describe these relationships, especially when these quantities are changing. For the special case of constant acceleration, these relationships simplify into a set of algebraic equations known as the equations of motion.

The core idea is to start from the definitions of velocity and acceleration as derivatives and then reverse the process using integration.

• Velocity is the rate of change of position with respect to time: $v = \frac{dx}{dt}$.
• Acceleration is the rate of change of velocity with respect to time: $a = \frac{dv}{dt}$.

Since acceleration is constant, $a$ is a fixed value, which makes the integration straightforward. We will integrate acceleration to find velocity, and then integrate velocity to find position.

Let's define our variables:

• $u$: initial velocity (velocity at time $t=0$)
• $v$: final velocity (velocity at time $t$)
• $a$: constant acceleration
• $t$: time elapsed
• $x_0$: initial position (position at time $t=0$)
• $x$: final position (position at time $t$)

1. Deriving the first equation of motion: $v = u + at$

   We begin with the definition of acceleration:

Since $a$ is constant, we can separate the variables and integrate both sides. We integrate velocity from its initial value $u$ to its final value $v$, and time from $0$ to $t$:

The integral of $dv$ is $v$, and the integral of $a \, dt$ (where $a$ is a constant) is $at$.

Applying the limits of integration:

Rearranging this gives us the first equation of motion:
> [!FORMULA]
> $$v = u + at$$
> This equation relates final velocity, initial velocity, constant acceleration, and time.

2. Deriving the second equation of motion: $x = x_0 + ut + \frac{1}{2}at^2$

Now we use the definition of velocity:

From the first equation, we know that $v = u + at$. Substituting this expression for $v$:

Again, we separate variables and integrate. We integrate position from its initial value $x_0$ to its final value $x$, and time from $0$ to $t$:

The integral of $dx$ is $x$. The integral of $(u + at) \, dt$ is $ut + \frac{1}{2}at^2$ (since $u$ and $a$ are constants).

Applying the limits of integration:

Rearranging this gives us the second equation of motion:
> [!FORMULA]
> $$x = x_0 + ut + \frac{1}{2}at^2$$
> This equation relates final position, initial position, initial velocity, constant acceleration, and time.
>
> > [!TIP]
> > If the initial position $x_0$ is taken as the origin (i.e., $x_0 = 0$), this equation simplifies to $x = ut + \frac{1}{2}at^2$. This is a common convention in many problems.

3. Deriving the third equation of motion: $v^2 = u^2 + 2a(x - x_0)$

This equation does not explicitly involve time. To derive it, we need a slightly different approach using the chain rule. …

Concept: Equations of Motion from the Velocity-Time Graph

Method: Graphical Derivation (slope and area, no integration)

The three equations of motion can be read directly off a velocity-time graph for constant acceleration — the slope gives the first equation, the area under the graph gives the second, and eliminating time between them gives the third. This is the classical geometric route, distinct from setting up and evaluating the integrals $\int dv$ and $\int dx$.

Setup

Draw the $v$-$t$ graph for an object with constant acceleration $a$: it is a straight line starting at $v=u$ (at $t=0$) and ending at $v$ (at time $t$).

Steps

1. First equation — from the slope. A straight line's slope is constant everywhere; here the slope of the $v$-$t$ line is the acceleration:

2. Second equation — from the area under the graph. Displacement equals the area swept out under the $v$-$t$ line between $0$ and $t$. That region is a trapezium with parallel sides $u$ and $v$, and height (width) $t$:

Substitute $v = u+at$ from Step 1:

$x-x_0 = \frac{1}{2}(u + u+at)\,t = ut + \frac{1}{2}at^2$ …