Equations of motion: Definition from Answers.com

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Equations of motion are equations that describe the behavior of a system (e.g., the motion of a particle under the influence of a force) as a function of time.^[1] Sometimes the term refers to the differential equations that the system satisfies (e.g., Newton's second law or Euler–Lagrange equations), and sometimes to the solutions to those equations.

1 Equations of uniformly accelerated linear motion
2 Classic version
- 2.1 Examples
- 2.2 Extension
3 Equations of circular motion
4 Derivation
5 See also
6 External links
7 References

Equations of uniformly accelerated linear motion

The equations that apply to bodies moving linearly (in one dimension) with constant acceleration are often referred to as "SUVAT" equations where the five variables are represented by those letters (s = displacement, u = initial velocity, v = final velocity, a = acceleration, t = time); the five letters may be shown in a different order.

The body is considered between two instants in time: one initial point and one current (or final) point. Problems in kinematics may deal with more than two instants, and several applications of the equations are then required. If a is constant, a differential, a dt, may be integrated over an interval from 0 to $Δ t$ ( $Δ t = t - t i$ ), to obtain a linear relationship for velocity. Integration of the velocity yields a quadratic relationship for position at the end of the interval.

$v = v_i + a \Delta t \,$

$s = s_i + v_i\Delta t + \tfrac{1}{2} a(\Delta t)^2 \,$

$s = s_i + \tfrac{1}{2} (v + v_i)\Delta t \,$

$v^2 = v_i^2 + 2a(s - s_i) \,$

where...

$v_i \,$ is the body's initial velocity

$s_i \,$ is the body's initial position

and its current state is described by:

$v \,$ , The velocity at the end of the interval

$s \,$ , the position at the end of the interval (displacement)

$\Delta t \,$ , the time interval between the initial and current states

$a \,$ , the constant acceleration, or in the case of bodies moving under the influence of gravity, g.

Note that each of the equations contains four of the five variables. Thus, in this situation it is sufficient to know three out of the five variables to calculate the remaining two.

Classic version

The above equations are often written in the following form:^[2]

$\begin{alignat}{3} & v && = u+at \qquad & \text{(1)} \\ & s && = \tfrac12(u+v)t \qquad & \text{(2)} \\ & s && = ut + \tfrac12 at^2 \qquad & \text{(3)} \\ & s && = vt - \tfrac12 at^2 \qquad & \text{(4)} \\ & v^2 && = u^2 + 2as \qquad & \text{(5)} \\ & a && = \frac{v-u}{t} \qquad & \text{(6)} \\ \end{alignat}$

By substituting (1) into (2), we can get (3), (4) and (5). (6) can be constructed by rearranging (1).

where

s = the distance between initial and final positions (displacement) (sometimes denoted R or x)

u = the initial velocity (speed in a given direction)

v = the final velocity

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 local acceleration of 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 fourth equation, we have:

$s= \frac{v^2 - u^2}{-2g}.$

Substituting and cancelling minus signs gives:

$s = \frac{u^2}{2g}.$

Extension

More complex versions of these equations can include a quantity Δs for the variation on displacement (s − s₀), s₀ for the initial position of the body, and v₀ for u for consistency.

$v = v_0 + at \,$

$s = s_0 + \tfrac{1}{2} (v_0 + v)t \,$

$s = s_0 + v_0 t + \tfrac{1}{2} at^2 \,$

$v^2 = v_0^2 + 2a\Delta s \,$

$s = s_0 + vt - \tfrac{1}{2} at^2 \,$

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

Equations of circular motion

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

$\omega=\omega_0+\alpha t\,$

$\phi=\tfrac12(\omega_0+\omega)t$

$\phi=\omega_0t+\tfrac12\alpha t^2\,$

$\omega^2=\omega_0^2+2\alpha\phi\,$

$\phi=\omega t-\tfrac12\alpha t^2\,$

where:

α

is the angular acceleration

ω

is the angular velocity

φ

is the angular displacement

ω 0

is the initial angular velocity.

Derivation

Motion equation 1

By definition of acceleration,

$\mathrm{ average\ acceleration } = \frac{\Delta v}{\Delta t}\quad\Rightarrow\quad\ a = \frac{v - u}{t}$

Hence

$at = v - u \,$

$v = u + at \,$

Motion equation 2

By definition,

$\mathrm{ average\ velocity } = \frac{s}{t}$

Hence

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

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

Motion equation 3

$t = \frac{v - u}{a}$

Using Motion Equation 2, replace t with above

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

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

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

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

Motion equation 4

Using Motion Equation 1 to replace u in Motion Equation 3 gives

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

External links

Equations of Motion Applet

References

^ Halliday, David; Resnick, Robert; Walker, Jearl (2004-06-16). Fundamentals of Physics (7 Sub ed.). Wiley. ISBN 0471232319.
^ Hanrahan, Val; Porkess, R (2003). Additional Mathematics for OCR. London: Hodder & Stoughton. p. 219. ISBN 0-340-86960-7.

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Equations of motion

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