Motion in a Straight Line

Beginner

A car pulls away from standing still; a cyclist brakes. Kinematics is the part of physics that describes such motion without asking what causes it. To describe motion along a straight line, we need three ideas.

Position tells you where the object is, measured from a starting point, in metres. Velocity tells you how fast the position changes and in which direction, in metres per second. Speed is just how fast, with no direction; velocity is speed plus a direction. Acceleration tells you how fast the velocity changes, in metres per second per second (m/s²).

If a cyclist rides 100 metres in 20 seconds, the average speed is $100÷20=5m/s$. If a car goes from standing still to 20 m/s in 4 seconds, it gains 5 m/s every second, so its acceleration is 5 m/s².

A common mix-up is to think that a fast object must be accelerating. A car cruising at a steady 30 m/s on a motorway has zero acceleration, because its velocity is not changing. A cyclist who brakes does have an acceleration: slowing down is also a change of velocity, and her acceleration points backward.

In the animation the car gains 6 m/s every second, and the grey dots mark where it is after each second. The gaps get bigger: that is what acceleration looks like.

Look at the gold gaps: 3 m, 9 m, 15 m, 21 m. They grow like the odd numbers 1, 3, 5, 7 (times 3 m). Galileo found this pattern in the early 1600s by timing balls rolling down a ramp, and it is the fingerprint of steady acceleration from rest.

Intermediate

What we measure, and from where

Two people can give different numbers for where the same car is at the same moment, and both be right. To describe a motion we first need a reference frame: an origin, a positive direction along the line and a clock. The position $x$ is then a signed number of metres: $x=30\mathrm{m}$ is 30 metres on the positive side of the origin, $x=-5\mathrm{m}$ is 5 metres on the other side. Another frame gives other numbers for the same motion.

The change of position is the displacement, $\Delta x=x-x_0$, where $x_0$ is the starting position. It has a sign and depends only on where the object started and where it ended. The distance travelled is the length of the whole path, and it is never negative. Walk 10 m forward and 10 m back: you have travelled 20 m, but your displacement is zero.

Velocity and acceleration are rates

Velocity measures how fast the position changes. Over a time interval $\Delta t$, the average velocity is
$v_{\mathrm{avg}}=\frac{\Delta x}{\Delta t}$

Its sign gives the direction of motion. Speed is the size of the velocity, without the sign. Average speed is the distance travelled divided by the time; it equals the size of $v_{\mathrm{avg}}$ only if the object never turns back. The velocity at one instant, the number on a speedometer, is the average velocity over a very short interval.

Acceleration measures how fast the velocity changes: $a=\Delta v/\Delta t$, in metres per second per second (m/s²). An acceleration of 6 m/s² adds 6 m/s to the velocity every second. When $v$ and $a$ have the same sign, the object speeds up; when their signs are opposite, it slows down. When $a=0$ the motion is uniform: $x=x_0+vt$, equal distances in equal times.

The constant-acceleration equations

If $a$ is constant, the velocity changes by the same amount every second, so its average over any interval lies exactly halfway between the starting and final values, $\frac{1}{2}(v_0+v)$. Multiplying that average by the time gives the displacement. These two facts lead to four equations:

• $v=v_0+at$
• $x=x_0+v_{0}t+\frac{1}{2}a{t}^2$
• $v^2=v_0^2+2a\Delta x$
• $\Delta x=\frac{1}{2}(v_0+v)t$

Each equation links four of the five quantities $\Delta x$, $v_0$, $v$, $a$ and $t$, so any three known values give the other two. Choose the equation that leaves out the quantity you neither know nor need: if time plays no part, use the third; if the acceleration is unknown, use the fourth.

Worked example: the car in the animation

The car starts from rest at the origin, so $v_0=0$ and $x_0=0$, and it accelerates at $a=6m/{\mathrm{s}}^2$. After $t=4\mathrm{s}$:
$v=v_0+at=0+6\cdot 4=24m/s$
$x=\frac{1}{2}a{t}^2=\frac{1}{2}\cdot 6\cdot 4^2=48\mathrm{m}$

These are the final values on the $v$ and $x$ readouts. The third equation checks them without using the time: $v^2=2\cdot 6\cdot 48=576{\mathrm{m}}^2/{\mathrm{s}}^2$, so $v=24m/s$. At 1, 2 and 3 s the car is at 3 m, 12 m and 27 m, so the gaps between the grey dots are 3, 9, 15 and 21 m, each 6 m longer than the one before. The average velocity is $48/4=12m/s$, half the final value, as the fourth equation predicts for a start from rest.

Meanwhile the blue $v$ arrow grows by the same step every second while the green $a$ arrow keeps its length: constant acceleration means equal gains in velocity in equal times, not equal distances.

Worked example: stopping distance

A driver at 14 m/s (about 50 km/h) sees a ball roll onto the road. She needs about 1.0 s to react, then brakes at a steady 7 m/s². How far does the car travel before it stops?

During the reaction time the car covers $14\cdot 1.0=14\mathrm{m}$ at constant speed. While braking, $v_0=14m/s$, $v=0$ and $a=-7m/{\mathrm{s}}^2$, negative because it points against the motion. The time is not asked for, so use the third equation:
$\Delta x=\frac{v^2-v_0^2}{2a}=\frac{0-14^2}{2\cdot (-7)}=14\mathrm{m}$

The car stops 28 m after the ball appears. At 28 m/s (about 100 km/h) the reaction distance doubles to 28 m, but the braking distance becomes $28^2/(2\cdot 7)=56\mathrm{m}$, four times as much: 84 m in all. Braking distance grows with the square of the speed, which is why speed limits near schools are so low.

Reading position–time and velocity–time graphs

On a position–time graph the slope is the velocity: a straight line means uniform motion, a horizontal line means rest, a downward slope means motion in the negative direction, and a curve that keeps getting steeper means speeding up.

On a velocity–time graph the slope is the acceleration, and the area under the line is the displacement (area below the time axis counts as negative). For the car, $v$ rises in a straight line from 0 to 24 m/s in 4 s: the slope is $24/4=6m/{\mathrm{s}}^2$, and the triangle under the line has area $\frac{1}{2}\cdot 4\cdot 24=48\mathrm{m}$. When $v_0$ is not zero the area is a trapezium, $\frac{1}{2}(v_0+v)t$, which is the fourth equation.

Common mistakes

• Speed is not velocity: two cars doing 20 m/s in opposite directions have the same speed but velocities of $+20$ and $-20m/s$.
• Zero velocity is not zero acceleration. A ball thrown straight up at 15 m/s stops after $15/9.81\approx 1.5\mathrm{s}$, about 11.5 m up, then falls back; at the top its acceleration is still $-g$.
• Negative acceleration does not always mean slowing down: a car speeding up in the negative direction has $v<0$ and $a<0$. Compare the signs of $v$ and $a$.
• The four equations need a constant $a$: split a motion that changes, such as cruising and then braking, into stages.
• Convert km/h to m/s first (divide by 3.6), and keep the ½ and the square in $\frac{1}{2}a{t}^2$: dropping them turns the car's 48 m into 96 m or 12 m.

Where you meet it

Stopping distances, a sprinter averaging 10 m/s over 100 m, lifts and trains that accelerate at about 1 m/s² to keep passengers comfortable: all of this is straight-line kinematics. It is also the vocabulary of the rest of mechanics. Free fall is this lesson with $a=-g$, projectile motion is two straight-line motions at once, and Newton's second law supplies the acceleration that these equations turn into a path.

Advanced

From average to instantaneous

The laws of motion speak about instants, yet any measured velocity covers an interval of time. How, then, can a velocity belong to a single instant? Take the average over a short interval, then see what it settles to as the interval shrinks.

Legend

$x$	Position of the car along its path; the curve is $x(t)$.	$\mathrm{m}$
$t$	The instant at which the velocity is wanted.	$\mathrm{s}$
$t+h$	A slightly later instant, a time $h$ after $t$.	$\mathrm{s}$
$h$	Length of the time interval from $t$ to $t+h$.	$\mathrm{s}$
$\Delta x$	Change of position from $t$ to $t+h$: the rise of the chord.	$\mathrm{m}$
$\bar{v}$	Average velocity over the interval: the slope of the chord, $\Delta x/h$.	$m/s$
$v$	Velocity at the instant $t$: the slope of the tangent, which the chords approach as $h\to 0$.	$m/s$

Figure 1 plots the position of the car in the animation, $x(t)=3t^2$, with $x$ in metres and $t$ in seconds. Pick an instant $t$ and a slightly later instant $t+h$, where $h$ is a short time. The straight line joining the two points of the curve is called a chord. Along it the position rises by $\Delta x$ while the time advances by $h$, so its slope $\Delta x/h$ is the average velocity $\bar{v}$ over the interval. Expanding $(t+h{)}^2=t^2+2th+h^2$ gives
$\bar{v}=\frac{\Delta x}{h}=\frac{3(t+h{)}^2-3t^2}{h}=\frac{6th+3h^2}{h}=6t+3h$

Now look at the dashed chord: its far end has slid down the curve towards the point at $t$. As $h$ shrinks, the chord turns and settles on the line labelled $v$, the tangent that just touches the curve at $t$. In the formula the term $3h$ goes to zero, so the slope of the chord tends to $6t$, the slope of the tangent. At $t=2\mathrm{s}$ the car is therefore doing exactly 12 m/s, even though every finite interval that starts at that instant gives a little more, $12+3h$.

This limit of the ratio is the derivative of $x$ with respect to $t$, and it defines the instantaneous velocity. The same step applied to the velocity defines the acceleration:
$v={\lim }_{\Delta t\to 0}\frac{\Delta x}{\Delta t}=\frac{dx}{dt},a=\frac{dv}{dt}=\frac{d^{2}x}{d{t}^2}$

Two consequences follow. First, no single snapshot can show a velocity, because a velocity compares positions at neighbouring instants. That answers Zeno's arrow paradox, in which the arrow seems motionless at every instant of its flight: at each instant it has a position, and its velocity lies in how that position changes nearby. Second, position, velocity and acceleration form a chain of derivatives, $x\to v\to a$. Newton's dot notation writes the two steps compactly as $\dot{x}$ and $\ddot{x}$.

Integrating back: constant acceleration

Going down the chain takes derivatives; going back up takes integrals. Knowing the acceleration, one integration gives the velocity and a second gives the position. The commonest case is a constant acceleration $a$, and every step of it can be read off a graph.

Legend

$v_0$	Velocity at the start.	$m/s$
$v$	Velocity after the time $t$: $v=v_0+at$.	$m/s$
$\bar{v}$	Mean velocity $\frac{1}{2}(v_0+v)$: the dashed level half-way between $v_0$ and $v$.	$m/s$
$at$	Velocity gained during $t$: the constant acceleration $a$ times the time.	$m/s$
$t$	Time elapsed since the start.	$\mathrm{s}$
$\Delta x$	Displacement during $t$: the shaded area under the line.	$\mathrm{m}$

Figure 2 plots velocity against time. The velocity starts at $v_0$. Because the acceleration is constant, the velocity grows by the same amount every second, so the graph is a straight line of slope $a$. After a time $t$ it has risen by $at$ (the bracket on the right), to $v=v_0+at$. As an integral, with the integration variable written $t^{\prime }$ to keep it apart from the upper limit $t$:
$v(t)=v_0+{\int }_0^{t}ad{t}^{\prime }=v_0+at$

Over a very short time the velocity is nearly constant, so the displacement is velocity times that time: a thin strip of area under the graph. Adding the strips, the displacement $\Delta x=x-x_0$ over the whole time $t$ is the shaded area under the line. Split it into a rectangle of height $v_0$ and width $t$, of area $v_{0}t$, and a triangle of base $t$ and height $at$, of area $\frac{1}{2}a{t}^2$:
$x(t)=x_0+{\int }_0^{t}v(t^{\prime })d{t}^{\prime }=x_0+v_{0}t+\frac{1}{2}a{t}^2$

The same area can also be counted in one piece. The dashed level $\bar{v}$ lies half-way between $v_0$ and $v$. Because the line is straight, it crosses that level exactly at the middle of the interval, at $t/2$. On the left half the line runs below the level, leaving a triangular gap; on the right half it runs above, adding a triangle. These are the two darker triangles, and they are equal, each with base $t/2$ and height $at/2$. Cut off the triangle on the right, drop it into the gap on the left, and the area under the line becomes a rectangle of height $\bar{v}$ and width $t$:
$\Delta x=\bar{v}t=\frac{1}{2}(v_0+v)t$

Each integration brought in a constant: $v_0$ the first time, $x_0$ the second. They are not decoration. The acceleration fixes the shape of the motion; the starting position and velocity pick out which motion of that shape actually happens. That is why Newton's second law, which supplies the acceleration, always needs exactly two pieces of starting information.

A further relation leaves out the time, which helps when the time is neither given nor asked for. It comes from the chain rule: since $dx/dt=v$,
$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx},\text{so}adx=vdv$
With $a$ constant, integrate the left side from $x_0$ to $x$ and the right side from $v_0$ to $v$. This gives $a(x-x_0)=\frac{1}{2}(v^2-v_0^2)$, that is
$v^2=v_0^2+2a\Delta x$

Multiply both sides of $a(x-x_0)=\frac{1}{2}(v^2-v_0^2)$ by the mass $m$ and it reads $ma\Delta x=\frac{1}{2}m{v}^2-\frac{1}{2}m{v}_0^2$. Since $ma$ is the force, this says that force times displacement equals the change of kinetic energy: the work–energy theorem, already hidden in kinematics.

Of the four constant-acceleration equations, $\Delta x=\frac{1}{2}(v_0+v)t$ is the oldest. Scholars at Merton College, Oxford, stated it around 1335 as the mean speed theorem, and Nicole Oresme soon proved it by comparing areas, much as Figure 2 does, some two and a half centuries before Galileo tested it with balls rolling down a ramp.

Galileo's odd-number rule follows from it too. Starting from rest, the distance covered during the $n$-th second is $x(n)-x(n-1)=\frac{1}{2}a[n^2-(n-1{)}^2]=\frac{1}{2}a(2n-1)$, so the distances in successive seconds grow as 1, 3, 5, 7. For the car, with $a=6m/{\mathrm{s}}^2$, they are 3, 9, 15 and 21 m.

Slopes and areas

Graphs of position, velocity and acceleration are the quickest way to read a motion, and every rule for reading them is a statement about derivatives (slopes) or integrals (areas). Figure 3 shows all the rules on one example.

Legend

$x$	Position; its graph (top) is a parabola.	$\mathrm{m}$
$v$	Velocity: the straight line of the middle graph, and the slope of the tangent to $x(t)$ at $t_1$.	$m/s$
$a$	Acceleration: constant here, so its graph (bottom) is a horizontal line.	$m/{\mathrm{s}}^2$
$t$	Time, the horizontal axis shared by all three graphs.	$\mathrm{s}$
$t_1$	The instant at which the tangent is drawn and the areas end.	$\mathrm{s}$
$\Delta x$	Change of position from 0 to $t_1$: the rise of $x(t)$, equal to the area under $v(t)$.	$\mathrm{m}$
$\Delta v$	Change of velocity from 0 to $t_1$: the rise of $v(t)$, equal to the area under $a(t)$.	$m/s$

Figure 3 stacks the three graphs of the car in the animation on one time axis: $x(t)=3t^2$ at the top, $v(t)=6t$ in the middle and the constant $a=6m/{\mathrm{s}}^2$ at the bottom. Read it first from top to bottom, by slopes. At the instant $t_1=4\mathrm{s}$ the tangent to the parabola $x(t)$ has a slope equal to the velocity there, $v=6\cdot 4=24m/s$: it climbs 24 m for every second. The middle graph $v(t)$ is a straight line because its slope, the acceleration, is the same at every instant.

The bending of $x(t)$ shows the sign of the acceleration, because $a$ is the rate at which the slope $v$ changes. A graph that curves upward has a growing slope, so $a>0$; one that curves downward has $a<0$; a straight stretch has $a=0$.

Now read from bottom to top, by areas. The area under $v(t)$ from 0 to $t_1$ is a triangle, $\frac{1}{2}\cdot 4\mathrm{s}\cdot 24m/s=48\mathrm{m}$, and it equals the rise $\Delta x$ of $x(t)$ over the same time: indeed $x(4)=3\cdot 16=48\mathrm{m}$. The area under $a(t)$ is a rectangle, $6m/{\mathrm{s}}^2\cdot 4\mathrm{s}=24m/s$, and it equals the rise $\Delta v$ of $v(t)$. In the figure each area carries the same label as the rise it produces.

Two cautions. First, area under $v(t)$ counts as negative where $v<0$, so it gives the displacement, not the distance travelled; the distance is $\int |v|dt$. Second, the graphs of $v$ and $a$ say nothing about where the motion started: shifting $x(t)$ up or down leaves both unchanged, so $x_0$ must always be given separately.

When the acceleration is not constant

Real accelerations change: an engine's push fades, air resists more at higher speed, a spring pulls harder the further it is stretched. The method depends on what the acceleration is a function of, and three cases cover most problems.

• Acceleration given as a function of time, $a(t)$: integrate twice, first $v=v_0+{\int }_0^{t}ad{t}^{\prime }$, then $x=x_0+{\int }_0^{t}vd{t}^{\prime }$. The worked problem below is of this kind.
• Acceleration that depends on velocity, $a(v)$, as with drag: since $a=dv/dt$, separate the variables as $dt=dv/a(v)$ and integrate. For the linear drag $a=-kv$, with $k$ a constant, this gives $v=v_0{e}^{-kt}$ and $x=x_0+(v_0/k)(1-e^{-kt})$. The body never quite stops, yet it covers only a finite distance, $v_0/k$. If the car of the animation, at 24 m/s, were slowed by such a drag alone with $k=0.5{\mathrm{s}}^{-1}$, it would coast at most $24/0.5=48\mathrm{m}$. For a body falling against quadratic drag, $a=g-b{v}^2$ with downward taken as positive, the acceleration dies away as $v$ approaches the terminal speed $\sqrt{g/b}$, at which $a=0$.
• Acceleration that depends on position, $a(x)$, as with springs: integrate $adx=vdv$ from above. For $a=-{\omega }^{2}x$ it gives $v^2=v_0^2-{\omega }^2(x^2-x_0^2)$. Multiplied by $\frac{1}{2}m$, this is energy conservation for an oscillator, found before energy has even been introduced.

Stepping forward numerically

Most accelerations met in practice lead to equations that no formula solves. The motion is then built up in small time steps, each using the acceleration known at its start. Figure 4 shows the simplest such scheme, Euler's method, on a case whose exact answer is known, so that its error can be seen.

Legend

$v$	Velocity of the coasting car; the blue curve is the exact $v(t)$, the broken line the numerical steps.	$m/s$
$t$	Time.	$\mathrm{s}$
$\Delta t$	Length of one time step.	$\mathrm{s}$
$a$	Acceleration $a=-kv$ found at the start of a step and kept for the whole step: the slope of the step (negative, since $v$ falls).	$m/{\mathrm{s}}^2$
$a\Delta t$	Change of velocity in one step: negative here, a decrease.	$m/s$

Figure 4 follows the coasting car with drag, $a=-kv$, with the numbers of the example above: $k=0.5{\mathrm{s}}^{-1}$ and $v_0=24m/s$. The blue curve is the exact solution $v(t)=v_0{e}^{-kt}$. The black broken line is the numerical estimate, with a deliberately coarse step $\Delta t=1\mathrm{s}$ so that the error is easy to see.

Every step follows the same rule. At the start of the step, work out the acceleration from the current velocity, $a=-kv$. Assume it keeps that value for the whole step, so the velocity changes by $a\Delta t$: replace $v$ by $v+a\Delta t$. Advance the position the same way, replacing $x$ by $x+v\Delta t$, and repeat from the new values. The highlighted step shows one pass: its slope is the $a$ found at its start, so across the time $\Delta t$ the velocity drops by $a\Delta t$ and no more.

Here each step multiplies the velocity by $1-k\Delta t=0.5$. After 1, 2, 3 and 4 s the steps give 12, 6, 3 and 1.5 m/s, against exact values of 14.6, 8.8, 5.4 and 3.2 m/s. The steps fall below the curve because the true deceleration weakens as $v$ falls during a step, while the numerical step keeps the stronger value it started with. The error is proportional to $\Delta t$: once the step is small, halving it roughly halves the error, so the steps converge to the exact motion as $\Delta t\to 0$. Orbits, weather forecasts and the physics of video games are computed with refined versions of this idea.

Worked problem: a fading push

Here is an acceleration that changes in time, handled by integrating twice. Suppose the car's push fades steadily, $a(t)=a_0(1-t/\tau )$, with $a_0=6m/{\mathrm{s}}^2$ and $\tau =8\mathrm{s}$, starting from rest at $x=0$. The acceleration falls in a straight line from $a_0$ to zero at $t=\tau$. How fast is the car going, and how far has it gone, when the push dies out?

Integrate once, with $v_0=0$:
$v(t)={\int }_0^t{a}_0\left(1-\frac{t^{\prime }}{\tau }\right)d{t}^{\prime }=a_0\left(t-\frac{t^2}{2\tau }\right)$
so $v(\tau )=a_0(\tau -\tau /2)=a_0\tau /2=24m/s$, the same final speed as in the animation. Integrate again, with $x_0=0$:
$x(t)=a_0\left(\frac{t^2}{2}-\frac{t^3}{6\tau }\right),x(\tau )=\frac{a_0{\tau }^2}{3}=\frac{6\cdot 64}{3}=128\mathrm{m}$

Compare the car in the animation, which reached 24 m/s after 4 s and 48 m: this one needs 8 s and 128 m. Its $v(t)$ is now a parabola that flattens out at $t=\tau$, where $a=0$, and the area under it is two thirds of the 8 s by 24 m/s rectangle, $\frac{2}{3}\cdot 192=128\mathrm{m}$.

A tempting shortcut fails. The average acceleration is $24/8=3m/{\mathrm{s}}^2$, but treating it as constant gives $\frac{1}{2}\cdot 3\cdot 8^2=96\mathrm{m}$, which is 32 m short. This car gains most of its speed early, so it spends longer going fast: its average speed is $128/8=16m/s$, not the 12 m/s of a steady build-up. The constant-acceleration formulas cannot be rescued by putting in an average acceleration.

Limits of validity

Everything above rests on three idealisations, and each one fails somewhere. First, the body is treated as a point with a single position. For a car or a planet that means following its centre of mass; spin and deformation need their own description.

Second, $x(t)$ is assumed smooth enough to be differentiated twice. A collision that seems to reverse a velocity at once is really a very large acceleration over a very short time. At the scale of atoms the idea of a sharp path $x(t)$ fails altogether, because quantum mechanics does not allow a particle a precise position and a precise velocity at the same time.

Third, Galilean kinematics assumes that all observers share one clock. Velocities then simply add: in a frame moving at constant velocity $u$, a body's velocity is $v^{\prime }=v-u$, while its acceleration is the same in every such frame. That is why Newton's laws hold in every inertial frame.

Special relativity shows that the shared clock is an approximation, valid for speeds far below the speed of light, $c\approx 3.00\times 10^{8}m/s$. For the car at 24 m/s the corrections are of order $v^2/c^2\approx 6\times 10^{-15}$, far below anything measurable. GPS satellites, at about 3.9 km/s, are another matter: this effect alone makes their clocks lose some 7 microseconds a day, and without a correction their position fixes would drift by kilometres within a day. Gravity, through general relativity, adds a larger effect in the opposite direction.

Where this leads

Kinematics describes motion; dynamics explains it. Newton's second law supplies the acceleration, $a=F/m$, and kinematics turns that acceleration into a trajectory: that is the whole programme of classical mechanics. The cases met here return throughout. Free fall is a constant acceleration of size $g$, pointing down. Projectile motion is two independent straight-line motions, one uniform and one uniformly accelerated. In circular motion the speed can stay constant while the velocity turns, which takes an acceleration $v^2/r$ towards the centre. Simple harmonic motion is $a=-{\omega }^{2}x$, drag supplies an $a(v)$, and $v^2=v_0^2+2a\Delta x$ returns as the work–energy theorem. Relative velocities and moving frames come next, in Vectors and Relative Motion.