Equation Of Motion

The equations that apply to bodies moving linearly (that is, one dimension) with uniform Acceleration are presented below.

LINEAR EQUATIONS OF MOTION

The body is considered at two instants in time: one "initial" point and one "current". Often, problems in kinematics deal with more than two instants, and several applications of the equations are required.

Note that each of the equations contains four of the five variables.
When using the above formulae, it is sufficient to know three out of the five variables to calculate remaining two.

CLASSIC VERSION

The above equations are often found in the following version:

:

v = u+at \,

s = rac {1} {2}(u+v) \cdot t

s = ut + rac {1} {2} a t^2

v^2 = u^2 + 2 a s \,

s = vt - rac {1} {2} a t^2

where

s

u

v

a

t

Examples

Many examples in kinematics involve Projectile s, for example a ball thrown upwards into the air.

Given initial speed ''u'', one can calculate how high the ball will travel before it begins to fall.

The acceleration is normal gravity ''g''. At this point one must remember that while these quantities appear to be Scalar s, the direction of displacement, speed and acceleration is important. They could in fact be considered as uni-directional vectors. Choosing ''s'' to measure up from the ground, the acceleration ''a'' must be in fact ''−g'', since the force of Gravity acts downwards and therefore also the acceleration on the ball due to it.

At the highest point, the ball will be at rest: therefore ''v'' = 0. Using the 4th equation, we have:

:

s= rac{v^2 - u^2}{-2g}

Substituting and cancelling minus signs gives:

:

s = rac{u^2}{2g}

Extension

More complex versions of these equations can include a quantity

\Delta

''s'' for the variation on displacement (''s'' - ''s''₀), ''s''₀ for the initial position of the body, and ''v''₀ for ''u'' for consistency.

:

v = v_0 + at \,

s = s_0 + \begin{matrix} rac{1}{2} \end{matrix} (v_0 + v)t \,

s = s_0 + v_0 t + \begin{matrix} rac{1}{2} \end{matrix}{at^2} \,

(v)^2 = (v_0)^2 + 2a \Delta s \,

s = s_0 + v t - \begin{matrix} rac{1}{2} \end{matrix}{at^2} \,

However a suitable choice of origin for the one-dimensional axis on which the body moves makes these more complex versions unnecessary.

ROTATIONAL EQUATIONS OF MOTION

The analogues of the above equations can be written for Rotation :

:

\omega = \omega_0 + \alpha t \,

\phi = \phi_0 
+ \begin{matrix} rac{1}{2} \end{matrix}(\omega_0 + \omega)t

\phi = \phi_0 + \omega_0 t + \begin{matrix} rac{1}{2} \end{matrix}\alpha {t^2} \,

(\omega)^2 = (\omega_0)^2 + 2\alpha \Delta \phi \,

\phi = \phi_0 + \omega t - \begin{matrix} rac{1}{2} \end{matrix}\alpha {t^2} \,

where:
:

\alpha

is the Angular Acceleration
:

\omega

is the Angular Velocity
:

\phi

is the Angular Displacement
:

\omega_0

is the initial angular velocity
:

\phi_0

is the initial angular displacement
:

\Delta \phi

is the variation on angular displacement (

\phi

\phi_0

).

DERIVATION

Motion equation 1

By definition of acceleration,
:

\ a = rac{v - u}{t}

Hence
:

at = v - u \,

v = u + at \,

Motion equation 2

By definition,
:

\mathrm{ average\ velocity } = rac{s}{t}

Hence
:

\begin{matrix} rac{1}{2} \end{matrix} (u + v) = rac{s}{t}

s = \begin{matrix} rac{1}{2} \end{matrix} (u + v)t

Motion equation 3

Insert ''Motion Equation 1'' into ''Motion Equation 2''
:

s = \begin{matrix} rac{1}{2} \end{matrix} (u + u + at)t

s = \begin{matrix} rac{1}{2} \end{matrix} (2u + at)t

s = ut + \begin{matrix} rac{1}{2} \end{matrix} at^2

Motion equation 4

:

t = rac{v - u}{a}

Using ''Motion Equation 2'', replace ''t'' with above
:

s = \begin{matrix} rac{1}{2} \end{matrix} (u + v) ( rac{v - u}{a} )

2as = (u + v)(v - u) \,

2as = v^2 - u^2 \,

v^2 = u^2 + 2as \,

Motion equation 5

Using ''Motion Equation 1'' to replace ''u'' in ''motion equation 3'' gives
:

s = vt - \begin{matrix} rac{1}{2} \end{matrix} at^2

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Equation Of Motion