Equations of Motion and Graphical Analysis

Introduction to Kinematic Equations

For an object moving in a straight line with constant acceleration, a set of fundamental equations, known as the kinematic equations of motion, can be used to relate displacement, velocity, acceleration, and time. These equations are essential for solving a wide range of problems in rectilinear motion.

Derivations of Kinematic Equations

Let's consider an object with initial velocity $v_0$ at time $t=0$, moving with constant acceleration $a$. After time $t$, its final velocity is $v$ and its displacement from the origin is $x$.

1. Derivation using Calculus

• Velocity-Time Relation ($v = v_0 + at$): From the definition of acceleration, $a = \frac{dv}{dt}$. Rearranging gives $dv = a\,dt$. Integrating both sides with limits from $v_0$ to $v$ and from 0 to $t$: $$\int_{v_0}^{v} dv = \int_{0}^{t} a \,dt = a \int_{0}^{t} dt$$ $$[v]_{v_0}^v = a[t]_0^t \implies v - v_0 = at \implies v = v_0 + at$$

• Position-Time Relation ($x = x_0 + v_0t + \frac{1}{2}at^2$): From the definition of velocity, $v = \frac{dx}{dt}$. We can substitute our first equation: $\frac{dx}{dt} = v_0 + at$. Rearranging gives $dx = (v_0 + at)\,dt$. Integrating with limits from initial position $x_0$ to final position $x$ and time from 0 to t: $$\int_{x_0}^{x} dx = \int_{0}^{t} (v_0 + at) \,dt$$ $$x - x_0 = [v_0t + \frac{at^2}{2}]_0^t \implies x - x_0 = v_0t + \frac{1}{2}at^2$$

• Position-Velocity Relation ($v^2 = v_0^2 + 2a(x - x_0)$): We use the chain rule: $a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}$. Rearranging gives $a\,dx = v\,dv$. Integrating with limits from $x_0$ to $x$ and from $v_0$ to $v$: $$\int_{x_0}^{x} a\,dx = \int_{v_0}^{v} v\,dv$$ `$$a(x-x_0) = [\frac{v^2}{2}]_{v_0}^v = \frac{v^2}{2} - \frac{v_0^2}{2} \implies v^2 = v_0^2 + 2a(x-x_0)$$

2. Derivation using v-t Graph

Consider the velocity-time graph for an object in uniformly accelerated motion. It's a straight line.

• Slope of v-t graph = Acceleration: $$a = \frac{v-v_0}{t-0} \implies v = v_0 + at$$
• Area under v-t graph = Displacement: The shape is a trapezoid. Area = $\frac{1}{2}(\text{sum of parallel sides}) \times \text{height}$. $$s = \frac{1}{2}(v_0 + v)t$$ Substituting $v=v_0+at$ into this gives $s = \frac{1}{2}(v_0 + v_0 + at)t = v_0t + \frac{1}{2}at^2$.

Example:

A ball is thrown vertically upward with a velocity of $20\ m/s$ from the top of a multistorey building. The height of the point from where the ball is thrown is 25.0 m from the ground. (Take $g = 10\ m/s^2$) (a) How high will the ball rise? (b) How long will it be before the ball hits the ground?

Solution: Let's set the origin (y=0) at the point of throw. The upward direction is positive. Initial velocity $v_0 = +20\ m/s$. Acceleration $a = -g = -10\ m/s^2$.

(a) Maximum Height: At the maximum height, the final velocity $v=0$. Using the kinematic equation without time, $v^2 = v_0^2 + 2as$: $0^2 = (20)^2 + 2(-10)(s)$ $0 = 400 - 20s$ $s = 20\ m$. So the ball rises 20 m above the top of the building.

(b) Time to Hit the Ground: The ground is at a final position of $s = -25\ m$ relative to our origin. Using the position-time equation $s = v_0t + \frac{1}{2}at^2$: $-25 = (20)t + \frac{1}{2}(-10)t^2$ $-25 = 20t - 5t^2$ Rearranging into a standard quadratic form: $5t^2 - 20t - 25 = 0$, which simplifies to $t^2 - 4t - 5 = 0$. Factoring gives $(t-5)(t+1)=0$. Since time cannot be negative, the answer is $t=5$ s.