Equations of Motion: Explanation, Derivation, and Applications
ANAND CLASSES study material and notes to learn the three equations of motion in physics with detailed derivations, explanations, conceptual questions, MCQs, and worksheets. Ideal for JEE, NEET, and CBSE Class 11 exam preparation.
The three equations of motion describe the relationship between an object’s displacement, velocity, acceleration, and time under uniform acceleration. These equations play a crucial role in solving problems in kinematics, particularly for competitive exams like JEE, NEET, and CBSE Class 11 Physics.
The three equations of motion are:
- First Equation : $\;v = u + at$
- Second Equation : $\;s = ut + \frac{1}{2} at^2$
- Third Equation : $\;v^2 – u^2 = 2as$
First Equation of Motion: Velocity-Time Relation
Statement: The final velocity of an object moving with uniform acceleration is given by: $$v = u + at$$
Derivation:
- Acceleration a is defined as the rate of change of velocity: $$a = \frac{dv}{dt}$$
- Rearranging: $$dv = a \:dt$$
- To determine how velocity changes over time, integrate both sides: $$\int_{u}^{v} dv = \int_{0}^{t} a \:dt$$
- Since acceleration a is constant, it comes out of the integral: $$\int_{u}^{v} dv = a \int_{0}^{t} dt$$
- Evaluating the integrals: $$v \:-\: u = a\: (t – 0)$$
- Simplifying: $$v = u + a\:t$$
Thus, we obtain the first equation of motion.
Second Equation of Motion: Displacement-Time Relation
Statement: The displacement ss of an object moving with uniform acceleration is given by: $s = ut + \frac{1}{2} at^2$
Derivation:
- Velocity is the rate of change of displacement: $$v = \frac{ds}{dt}$$
- Substituting $$v = u + at$$ $$\frac{ds}{dt} = u + at$$
- Rearranging: $$ds = (u + at) dt$$
- Integrating both sides from s = 0 to s and t = 0 to t : $$\int_{0}^{s} ds = \int_{0}^{t} (u + at) dt$$
- Splitting the integral: $$\int_{0}^{s} ds = \int_{0}^{t} u dt + \int_{0}^{t} at dt$$
- Evaluating the first integral : $$\int_{0}^{t} u dt = u \int_{0}^{t} dt = u [t]_0^t = ut$$
- Evaluating the second integral : $$\int_{0}^{t} at dt = a \int_{0}^{t} t dt = a \left[ \frac{t^2}{2} \right]_0^t = \frac{1}{2} at^2$$
- Adding both results : $$s = ut + \frac{1}{2} at^2$$
Thus, we obtain the second equation of motion.
Third Equation of Motion: Velocity-Displacement Relation
Statement: The final velocity vv of an object moving with uniform acceleration is given by : $v^2 – u^2 = 2as$
Derivation:
- Using the chain rule for acceleration: $$a = \frac{dv}{dt}$$
- Multiplying both sides by dx/dx : $$a = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$$
- Rearranging: $$v dv = a dx$$
- Integrating both sides from u to v and x = 0 to x = s : $$\int_{u}^{v} v dv = \int_{0}^{s} a dx$$
- Evaluating the left integral : $$\int_{u}^{v} v dv = \left[ \frac{v^2}{2} \right]_{u}^{v} = \frac{v^2}{2} – \frac{u^2}{2}$$
- Evaluating the right integral : $$\int_{0}^{s} a dx = a [x]_0^s = a (s – 0) = as$$
- Equating both results: $$\frac{v^2}{2} – \frac{u^2}{2} = as$$
- Multiplying both sides by 2: $$v^2 – u^2 = 2as$$
Thus, we obtain the third equation of motion.
Conceptual Questions and Answers
- What happens if acceleration is zero in the equations of motion?
- If a = 0, then v = u and s = ut, meaning the object moves with constant velocity.
- Why do we use integration in the derivation?
- Integration helps us sum up small changes in velocity and displacement over time.
Quick Revision Points
- First equation: $v = u + at $(Velocity-Time relation)
- Second equation: $s = ut + \frac{1}{2} at^2$ (Displacement-Time relation)
- Third equation: $v^2 – u^2 = 2as$ (Velocity-Displacement relation)
- These equations apply only to uniformly accelerated motion.
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