Introduction
One of the most fascinating aspects of physics is its ability to describe motion with just a few elegant mathematical relationships. Among these, the equations of motion hold a special place. They allow us to predict where an object will be, how fast it will move, and how long it will take to reach a certain point, provided the object is under uniform acceleration.
From a car speeding up on a straight road to a ball falling freely under gravity, uniformly accelerated motion governs countless real-world phenomena. These equations not only form the foundation of classical mechanics but also serve as stepping stones to advanced topics like relativity and astrophysics.
In this article, we’ll explore uniformly accelerated motion in depth, understand the concept of acceleration, and derive the three famous equations of motion step by step. We’ll also highlight their applications, historical context, and significance in both daily life and scientific exploration.
What is Uniformly Accelerated Motion?
Uniformly accelerated motion is motion in which the acceleration of the body remains constant over time.
- Acceleration (aaa) = rate of change of velocity with respect to time.
Mathematically: a=ΔvΔt=v−uta = \frac{\Delta v}{\Delta t} = \frac{v – u}{t}a=ΔtΔv=tv−u
where:
- uuu = initial velocity
- vvv = final velocity
- aaa = constant acceleration
- ttt = time
Because the acceleration remains constant, the velocity changes uniformly, and this makes the motion predictable with three fundamental equations.
The Three Equations of Motion
- First equation: v=u+atv = u + atv=u+at (Relates velocity, acceleration, and time)
- Second equation: s=ut+12at2s = ut + \frac{1}{2}at^2s=ut+21at2 (Relates displacement, time, and acceleration)
- Third equation: v2=u2+2asv^2 = u^2 + 2asv2=u2+2as (Relates velocity and displacement without time)
Each of these can be derived in multiple ways—algebraically, graphically, and using calculus. Let’s explore them step by step.
Derivation of the First Equation of Motion (v=u+atv = u + atv=u+at)
Method 1: Algebraic Definition
Acceleration is defined as the change in velocity per unit time: a=v−uta = \frac{v – u}{t}a=tv−u
Rearranging: v=u+atv = u + atv=u+at
That’s our first equation of motion.
Method 2: Graphical Approach
- Draw a velocity-time graph with initial velocity uuu.
- After time ttt, the velocity becomes vvv.
- The slope of the velocity-time graph represents acceleration.
Since slope = rise/run = (v−u)/t=a(v – u)/t = a(v−u)/t=a, we rearrange to get: v=u+atv = u + atv=u+at
Derivation of the Second Equation of Motion (s=ut+12at2s = ut + \tfrac{1}{2}at^2s=ut+21at2)
Method 1: Using Average Velocity
Displacement (sss) = average velocity × time.
Average velocity = (u+v)/2(u + v)/2(u+v)/2.
So, s=(u+v)2×ts = \frac{(u+v)}{2} \times ts=2(u+v)×t
From the first equation, v=u+atv = u + atv=u+at. Substituting: s=(u+(u+at))2×ts = \frac{(u + (u+at))}{2} \times ts=2(u+(u+at))×t s=(2u+at)2×ts = \frac{(2u + at)}{2} \times ts=2(2u+at)×t s=ut+12at2s = ut + \tfrac{1}{2}at^2s=ut+21at2
Method 2: Graphical Approach
- In a velocity-time graph, displacement is the area under the graph.
- The graph is a trapezium with parallel sides uuu and vvv and base ttt.
- Area = 12(u+v)×t\tfrac{1}{2}(u+v) \times t21(u+v)×t.
- Substituting v=u+atv = u + atv=u+at, we again get: s=ut+12at2s = ut + \tfrac{1}{2}at^2s=ut+21at2
Derivation of the Third Equation of Motion (v2=u2+2asv^2 = u^2 + 2asv2=u2+2as)
Method 1: Elimination of Time
From the first equation: v=u+at⇒t=v−uav = u + at \quad \Rightarrow \quad t = \frac{v-u}{a}v=u+at⇒t=av−u
From the second equation: s=ut+12at2s = ut + \tfrac{1}{2}at^2s=ut+21at2
Substitute t=(v−u)/at = (v-u)/at=(v−u)/a: s=u⋅v−ua+12a(v−ua)2s = u \cdot \frac{v-u}{a} + \tfrac{1}{2}a \left(\frac{v-u}{a}\right)^2s=u⋅av−u+21a(av−u)2
Simplify: s=u(v−u)a+(v−u)22as = \frac{u(v-u)}{a} + \frac{(v-u)^2}{2a}s=au(v−u)+2a(v−u)2
Multiply throughout by 2a2a2a: 2as=2u(v−u)+(v−u)22as = 2u(v-u) + (v-u)^22as=2u(v−u)+(v−u)2
Expand: 2as=2uv−2u2+v2−2uv+u22as = 2uv – 2u^2 + v^2 – 2uv + u^22as=2uv−2u2+v2−2uv+u2 2as=v2−u22as = v^2 – u^22as=v2−u2
So, v2=u2+2asv^2 = u^2 + 2asv2=u2+2as
Method 2: Graphical Approach
The displacement is the area under the velocity-time graph. Using geometry and the relation between slope and acceleration, the same result emerges.
Alternative Derivations Using Calculus
- First equation: a=dvdt⇒dv=a dta = \frac{dv}{dt} \quad \Rightarrow \quad dv = a \, dta=dtdv⇒dv=adt Integrating: ∫dv=∫a dt⇒v=u+at\int dv = \int a \, dt \quad \Rightarrow \quad v = u + at∫dv=∫adt⇒v=u+at
- Second equation: v=dsdt=u+atv = \frac{ds}{dt} = u + atv=dtds=u+at Integrating: s=∫(u+at) dt=ut+12at2s = \int (u+at) \, dt = ut + \tfrac{1}{2}at^2s=∫(u+at)dt=ut+21at2
- Third equation: a=dvdt=dvdsdsdt=vdvdsa = \frac{dv}{dt} = \frac{dv}{ds}\frac{ds}{dt} = v \frac{dv}{ds}a=dtdv=dsdvdtds=vdsdv a ds=v dva \, ds = v \, dvads=vdv Integrating: ∫a ds=∫v dv\int a \, ds = \int v \, dv∫ads=∫vdv as=12(v2−u2)as = \tfrac{1}{2}(v^2 – u^2)as=21(v2−u2) v2=u2+2asv^2 = u^2 + 2asv2=u2+2as
Real-Life Examples of Uniformly Accelerated Motion
- Free fall under gravity:
- Object dropped from rest: u=0u = 0u=0, a=ga = ga=g.
- Equations simplify to: v=gt,s=12gt2,v2=2gsv = gt, \quad s = \tfrac{1}{2}gt^2, \quad v^2 = 2gsv=gt,s=21gt2,v2=2gs
- Vehicles:
- A car accelerating from rest with constant acceleration follows these equations.
- Sports:
- Sprinters, cyclists, and swimmers accelerating from start blocks experience uniformly accelerated motion.
- Spacecraft:
- Rockets accelerating under thrust in space can often be modeled using these equations.
Applications in Science and Engineering
- Physics: Core principles in kinematics and dynamics.
- Engineering: Used in motion analysis of machines, vehicles, and structures.
- Astronomy: To calculate planetary motion approximations.
- Aviation: Predicting aircraft acceleration on runways.
- Safety: Car crash analysis relies on these equations to study impact forces.
Common Misconceptions
- “Acceleration means speed only increases.”
- Not always. Negative acceleration (deceleration) decreases velocity.
- “Equations work for any acceleration.”
- No, they only hold when acceleration is constant.
- “Displacement is always positive.”
- Displacement can be negative depending on direction.
Historical Perspective
- The foundations of uniformly accelerated motion were laid by Galileo Galilei in the 16th century.
- He studied falling bodies and inclined planes, concluding that acceleration under gravity is uniform.
- Newton later formalized these principles into his laws of motion.
Leave a Reply