Constant-Acceleration Motion

01Constant-Acceleration Motion

SUVAT Formulae & Derivation

State and derive the equations for motion with constant acceleration.

Five quantities, one constant acceleration

For one interval, let $s$ be displacement, $u$ initial velocity, $v$ final velocity, $a$ constant acceleration and $t$ elapsed time. Use a single positive direction for $s,u,v,a$ .

Equation	Quantity absent
$v=u+at$	$s$
$s=\tfrac12(u+v)t$	$a$
$s=ut+\tfrac12at^2$	$v$
$s=vt-\tfrac12at^2$	$u$
$v^2=u^2+2as$	$t$

Use gradient and signed area

A constant-acceleration interval, shown for

v>u>0

. The shaded area is displacement.

The velocity–time graph is straight because $a$ is constant. Its gradient gives $a=(v-u)/t$ , hence $v=u+at$ . Its signed area gives $s=\tfrac12(u+v)t$ . Substituting $v=u+at$ gives $s=ut+\tfrac12at^2$ ; substituting $u=v-at$ gives $s=vt-\tfrac12at^2$ .

For the time-free equation, combine $2s=(u+v)t$ with $at=v-u$ : $2as=(u+v)(v-u)=v^2-u^2.$ This form avoids dividing by $a$ , so it also accommodates $a=0$ .

Alternatively, integrating the constant acceleration gives $v=at+C$ ; the initial condition $v(0)=u$ gives $C=u$ . Integrating again, with displacement zero at $t=0$ , gives $s=ut+\tfrac12at^2$ .

02Constant-Acceleration Motion

Using SUVAT

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List the known and unknown quantities

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Use two conditions for two unknowns

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03Constant-Acceleration Motion

Multistage Motion

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Start a fresh interval when acceleration changes

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04Constant-Acceleration Motion

Vertical Motion

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Gravity keeps the same direction throughout

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05Constant-Acceleration Motion

Maximum Height & Flight Time

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At the top, velocity is zero

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