Equation of motion

From Academic Kids

In advanced physics, equations of motion usually refer to the Euler-Lagrange equations, differential equations derived from the Lagrangian. The following article is about elementary physics only.

Contents

Linear equations of motion

In kinematics, four equations of motion (or kinematic equations) apply to bodies moving linearly (that is, one dimension) with uniform acceleration.

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.

The body's initial speed is denoted <math>v_i \,<math>. Its current state is described by:

<math>d \,<math>, the distance travelled from initial state
<math>v_f \,<math>, the current speed
<math>\Delta t \,<math>, the time between the initial and current states

The constant acceleration is denoted a, or in the case of bodies moving under the influence of gravity, g.

<math>v_f = v_i + a\Delta t \,<math>
<math>d = \begin{matrix} \frac{1}{2} \end{matrix} (v_i + v_f)\Delta t<math>
<math>d = v_i\Delta t + \begin{matrix} \frac{1}{2} \end{matrix} a\Delta t^2<math>
<math>v_f^2 = v_i^2 + 2ad \,<math>

Classic version

The above equations are often found in the following version:

<math>v = u+at \,<math>
<math>s = ( \frac{u+v} {2}) \cdot t <math>
<math>s = ut + \frac {1} {2} a t^2 <math>
<math>v^2 = u^2 + 2 a s \,<math>

where

s = the distance travelled from the initial state to the final state
u = the initial speed
v = the final speed
a = the constant acceleration
t = the time taken to move from the initial state to the final state

Examples

Many examples in kinematics involve projectiles, 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 scalars, 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:

<math>s= \frac{v^2 - u^2}{-2g}<math>

Substituting and cancelling minus signs gives:

<math>s = \frac{u^2}{2g}<math>

Extension

More complex versions of these equations can include a quantity <math>\Delta<math>s for the variation on displacement (s - s0), s0 for the initial position of the body, and v0 for u for consistency.

<math>v = v_0 + at \,<math>
<math>s = s_0 + \begin{matrix} \frac{1}{2} \end{matrix} (v_0 + v)t \,<math>
<math>s = s_0 + v_0 t

+ \begin{matrix} \frac{1}{2} \end{matrix}{at^2} \,<math>

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

Note, however, that a suitable choice of origin for the one-dimensional axis on which the body moves renders this complication superfluous.

Rotational equations of motion

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

<math> \omega = \omega_0 + \alpha t \,<math>
<math> \phi = \phi_0

+ \begin{matrix} \frac{1}{2} \end{matrix}(\omega_0 + \omega)t <math>

<math> \phi = \phi_0 + \omega_0 t

+ \begin{matrix} \frac{1}{2} \end{matrix}\alpha {t^2} \,<math>

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

Here, <math>\alpha<math> is the angular acceleration, <math>\omega<math> is the angular velocity and <math>\phi<math> is the angular displacement; <math>\omega_0<math> is the initial angular velocity, <math>\Delta \phi<math> is the variation on angular displacement (<math>\Delta \phi<math> - <math>\phi<math>) and <math>\phi_0<math> is the initial angular displacement.

Derivation

All of the above Motion equations are derived from two well-known equations, these are:

Equation 1

<math>\ a = \frac{v - u}{t}<math>

Equation 2

<math> \mathrm{ average\ velocity } = \frac{s}{t}<math>

Motion equation 1

Using Equation 1

<math>at = v - u \,<math>
<math>v = u + at \,<math>

Motion equation 2

Using Equation 2

<math> \begin{matrix} \frac{1}{2} \end{matrix} (u + v) = \frac{s}{t}<math>
<math>s = \begin{matrix} \frac{1}{2} \end{matrix} (u + v)t<math>

Motion equation 3

Insert Motion Equation 1 into Motion Equation 2

<math>s = \begin{matrix} \frac{1}{2} \end{matrix} (u + u + at)t<math>
<math>s = \begin{matrix} \frac{1}{2} \end{matrix} (2u + at)t<math>
<math>s = ut + \begin{matrix} \frac{1}{2} \end{matrix} at^2<math>

Motion equation 4

<math>t = \frac{v - u}{a}<math>

Using Motion Equation 2, replace t with above

<math>s = \begin{matrix} \frac{1}{2} \end{matrix} (u + v) ( \frac{v - u}{a} )<math>
<math>2as = (u + v)(v - u) \,<math>
<math>2as = v^2 - u^2 \,<math>
<math>v^2 = u^2 + 2as \,<math>

See also

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