Q3 - Constant-acceleration equations

You already know

Velocity–time gradient gives acceleration, and signed area gives displacement. A trapezium has area equal to half the sum of its parallel sides multiplied by their separation. The identity $(v-u)(v+u)=v^2-u^2$ is the difference of two squares.

The five motion quantities

Over one interval, $u$ is initial velocity, $v$ is final velocity, $a$ is constant acceleration, $t$ is elapsed time and $s$ is displacement. In straight-line motion $u,v,a,s$ are signed along a chosen axis; elapsed time satisfies $t\ge0$. Initial position need not be zero: final position is initial position plus $s$.

Worked example 1: deriving velocity and displacement

Question. A particle has constant acceleration $a$ along a line. Derive $v=u+at$ and $s=ut+\frac12at^2$ from its velocity–time graph.

Plan. Use the graph's gradient, then use its signed trapezium area and eliminate $v$.

Why this step? Replacing final velocity by $u+at$ expresses displacement using the initial velocity and the constant acceleration.

Answer. The two equations follow from the graph's gradient and area. Both also hold at $t=0$, when $v=u$ and $s=0$.

Check. For $u=4$, $a=2$, $t=3$, they give $v=10$ and $s=12+9=21$, agreeing with the concrete graph calculation.

Worked example 2: deriving an equation without time

Question. Starting from $v-u=at$ and $2s=(u+v)t$, derive $v^2=u^2+2as$, including the case $a=0$.

Plan. Multiply the velocity difference by the velocity sum, then replace $(u+v)t$ by $2s$.

Why this step? Substitution removes $t$ without dividing by $a$, which might be zero.

Answer. $v^2=u^2+2as$, valid for constant acceleration including $a=0$.

Check. With $u=4,v=10,a=2,s=21$, both sides equal $100$. If $a=0$, the original velocity relation gives $v=u$, consistent with the result.

Constant acceleration along a line

For one interval with constant acceleration,

Here $s$ is displacement, not automatically distance.
In the formula booklet.

Watch out

Wrong: “Every journey has $s=\frac12(u+v)t$.” Its endpoint average equals average velocity under the constant-acceleration condition, not for an arbitrary velocity curve. Right: establish constant acceleration before using these equations.

Worked example 3: braking until the first stop

Question. A car travels at $18\,\mathrm{m\,s^{-1}}$ and has constant acceleration $-3\,\mathrm{m\,s^{-2}}$ until it first stops. Taking forward as positive, find the stopping time and distance.

Plan. Set $v=0$, find time from the velocity equation, then find displacement without time.

Why this step? The acceleration is negative along the chosen axis; “braking” does not make its signed value positive.

Answer. The car first stops after $6\,\mathrm s$, having travelled $54\,\mathrm m$.

Check. Its velocity remains nonnegative before the stop, so distance equals displacement. The triangular graph area is $\frac12(6)(18)=54\,\mathrm m$.

Worked example 4: an interval containing a reversal

Question. A particle has initial velocity $-5\,\mathrm{m\,s^{-1}}$ and constant acceleration $2\,\mathrm{m\,s^{-2}}$ for $4\,\mathrm s$. Find its final velocity, displacement and distance travelled.

Plan. Use the supplied equations for velocity and displacement, then split the distance where velocity is zero.

Why this step? The final displacement combines two opposite directions; distance needs both travelled legs separately.

Answer. Final velocity is $3\,\mathrm{m\,s^{-1}}$, displacement is $-4\,\mathrm m$, and distance is $8.5\,\mathrm m$.

Check. The velocity graph has two triangles: $\frac12(2.5)(5)=6.25$ and $\frac12(1.5)(3)=2.25$ metres. Their difference is $-4$ and their sum is $8.5$.

Watch out

Wrong: extend example 3 beyond $t=6$ and conclude that the braking car must reverse. The model was stated only until its first stop. Right: end that stage at $v=0$; further motion needs further information. A particle explicitly given constant acceleration through a reversal, as in example 4, is a different interval.