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Posts Tagged ‘Equations of motion’

Anyone who has studied mechanics / dynamics will have come across Newton’s equations of motion (not to be confused with his laws of motion). The ones I get my students to use are


  1. v = u + at \; \; \text{  (Equ. 1)}
  2. s = ut + \frac{1}{2} at^{2} \; \; \text{  (Equ. 2)}
  3. v^{2} = u^{2} + 2 as \; \; \text{  (Equ. 3)}


where u is the initial velocity, v is the velocity at time t, s is the displacement and a is the acceleration. Note, these equations are only true for constant acceleration, but that actually covers quite a lot of situations. They can all be derived from the definition of acceleration.



Newton's equations of motion can  be derived from his 2nd law  of motion.

Newton’s equations of motion can be derived from his 2nd law of motion.



Derivation of Equation 1

We start off with out definition of acceleration, which is the rate of change of velocity. Writing that mathematically,



a = \frac{dv}{dt}

This is an example of a first order differential equation. To solve it we integrate. So we have


a \int dt = \int dv


When we integrate without limits, we have to include a constant term, so we can write


at = v + C


where C is our constant. To determine the value of the constant we need to put in some conditions, such as (but not necessarily) initial conditions. When t=0 we have defined that v=u, so we can write


0 = u + C \rightarrow C=-u \rightarrow \boxed{v = u + at \; \; \text{ (Equ. 1)} }

Derivation of Equation 2

To derive equation two, which we notice involves the displacement (the vector equivalent of distance), we do the following



v = \frac{ds}{dt} = u + at \rightarrow ds = \int u dt + \int at dt
s = ut + \frac{at^{2}}{2} + C


When t=0 \text{ then } s=0 so we can write


0 = 0 + 0 +C \rightarrow C=0 \rightarrow \boxed{s = ut + \frac{1}{2}at^{2} \; \; \text{ (Equ. 2)} }

Derivation of Equation 3

To derive equation 3 we use the trick of writing the acceleration a in terms of the velocity v and the displacement s. To do this we write



a = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = \frac{dv}{ds} \cdot v = v \frac{dv}{ds}


So, writing


v \frac{dv}{ds} = a \rightarrow \int v dv = \int a ds \rightarrow \frac{v^{2}}{2} = as + C
Again, we can work out C by remembering that s=0 \text{ when } t=0 so


\frac{u^{2}}{2} = 0 + C \rightarrow C = \frac{u^{2}}{2}


and so


\frac{ v^{2} }{2} = as + \frac{ u^{2} }{2} \rightarrow \boxed{ v^{2} = u^{2} + 2 as \; \; \text{ (Equ. 3)} }

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