Describing Motion

A cup on your desk is perfectly still. Seen from the Moon it is racing around with the spinning Earth. Both statements are true, and that is the first idea of kinematics.

At rest compared to what?#

The everyday definition seems obvious: something is at rest if its position does not change as time passes, and moving if it does. But “position” always means position relative to something. A book on a table keeps its place relative to the room. To an astronaut standing on the Moon, the room, the table and the book are all sweeping through space together.

So rest and motion are not properties of the object alone. They describe a relationship between the object and an observer. Nothing is absolutely at rest or absolutely moving: the Moon moves relative to the book, and the book moves relative to the Moon.

Frames of reference#

To pin down where a particle is, we need something to measure from. The usual choice is three mutually perpendicular axes, X, Y and Z, plus a clock to keep time. The particle’s coordinates $(x, y, z)$ then locate it, and the clock tells us when.

There is no rule about which frame to pick. Choose whatever makes the description easiest. Sitting in a train, it is natural to attach the axes to your compartment. Then a suitcase on the luggage rack has constant coordinates (unless the train jerks) and is at rest, while stations, poles and trees change their coordinates and are moving. That is exactly why passengers say things like “Lonavala is coming” or “we’ve passed Pune.”

Often the frame is clear from context and nobody bothers to mention it. “The bus is moving but the rickshaw has stopped” obviously means relative to the road. From here on, unless stated, we will assume a suitable frame has been chosen and describe motion within it.

Your fitness tracker and a map app answer two different questions. One asks how far you walked; the other asks how far you are from where you started.

Two ways to measure a trip#

Suppose a particle is at point A at time $t_1$ and at point B at time $t_2$, travelling between them along some winding path. Two quantities describe this trip:

• The distance travelled is the length of the actual path from A to B.
• The displacement is the straight arrow from the starting point A to the finishing point B. Its size is the straight-line separation, and its direction is from start to finish.

Displacement has a magnitude and a direction. Does it also add by the triangle rule? Yes. Slide a coin across a table while someone pushes the table across the room, and the coin’s net displacement relative to the room is the vector sum of the two displacements. So displacement is a vector. Distance only has a size, so it is a scalar.

“We averaged 60 km/h” summarises a whole road trip in one number. The speedometer needle tells a different, moment-by-moment story. Both are speed, measured over different stretches of time.

Average speed#

If a particle covers a distance $s$ between times $t_1$ and $t_2$, its average speed over that interval is

Average speed captures the overall quickness of the trip. Think of a 300 km drive that takes 5 hours: you averaged 60 km/h, though you probably crawled through towns, sped on the highway and stopped for chai. The average hides all that detail.

Instantaneous speed#

A sprinter leaves the blocks slowly and builds up. To capture how fast she is going at one particular moment $t$, take a short interval from $t$ to $t + \Delta t$ in which she covers $\Delta s$, compute $\Delta s/\Delta t$, and then let $\Delta t$ shrink toward zero. Both $\Delta s$ and $\Delta t$ vanish, but their ratio approaches a definite limit, just as in the lesson on derivatives:

Average speed belongs to an interval; instantaneous speed belongs to an instant. When people just say “speed,” they mean the instantaneous kind.

Going backwards: distance from speed#

If the speed at time $t$ is $v$, then in a tiny interval $dt$ the particle covers $ds = v\,dt$. Add up all those little pieces from $t_1$ to $t_2$:

On a speed–time graph that integral is simply the area under the curve between the two times.

The dimension of speed is $[LT^{-1}]$ and its SI unit is the metre per second (m/s).

Speed tells you how fast. Velocity tells you how fast and which way, and it is built on displacement, not distance.

Average velocity#

If a particle is at A at time $t_1$ and at B at time $t_2$, its displacement over that interval is the vector $\vec{AB}$. Divide by the elapsed time:

Dividing a vector by a positive number keeps its direction, so average velocity points from A to B. It is a vector.

Instantaneous velocity#

As with speed, we zoom in. Take the displacement $\Delta\vec r$ in a short interval $\Delta t$ after time $t$ and let the interval shrink:

In a vanishingly short interval the displacement points along the path itself, so velocity is always tangent to the trajectory. And over such a short stretch the chord and the arc are the same length, $|d\vec r| = ds$, so

The magnitude of the velocity is the speed. Unlike the averages, which can be very different, the instantaneous speed and the size of the instantaneous velocity always agree.

Press the accelerator, hit the brakes, or turn the steering wheel. Each one changes your velocity, and so each one is an acceleration.

When velocity changes#

A particle whose velocity stays the same as time passes is in uniform motion. If the velocity changes, the particle is accelerating. Velocity is a vector, so it can change in three ways: in size, in direction, or in both.

If the velocity is $\vec v_1$ at time $t_1$ and $\vec v_2$ at $t_2$, the average acceleration over that interval is

Like average velocity, it uses only the two end values; how the velocity varied in between does not enter. Shrink the interval to get the instantaneous acceleration, usually just called the acceleration:

Signs on a straight line#

On a straight line, choose it as the x-axis. Position is a single coordinate $x$, and

A positive $v$ means moving toward +x; negative means toward −x. The same goes for $a$. Whether the particle speeds up or slows down depends on how the two signs compare:

• $v$ and $a$ with the same sign → speed increases.
• $v$ and $a$ with opposite signs → speed decreases. We call this deceleration.

Deceleration is just acceleration pointing against the motion. An acceleration of 2.0 m/s² toward the east is exactly the same thing as a deceleration of 2.0 m/s² toward the west.

If the acceleration never changes, the whole future of a straight-line motion is fixed by just two numbers: where it starts moving from and how fast. Three short equations do all the work.

Setting up#

Put the line of motion along the x-axis, start the clock at $t = 0$ and, usually, put the origin where the particle is at that moment. Let the acceleration $a$ be constant, the velocity at $t = 0$ be $u$, and the velocity at time $t$ be $v$.

Deriving the equations#

Signs carry direction: $u$, $v$, $a$ and $x$ are each positive or negative according to whether they point along +x or −x.

The most familiar constant acceleration of all is right under your feet. Drop anything, and gravity applies the same equations of motion every single time.

Falling is constant acceleration#

Ignore air resistance and let go of an object near the Earth’s surface. It falls along a vertical straight line with an acceleration that points straight down and is very nearly constant, as long as the height involved is tiny compared with the Earth’s radius (about 6400 km). Its magnitude is about 9.8 m/s² (or 32 ft/s²), and it gets its own symbol, $g$.

Because $g$ is constant, the three equations from the previous lesson apply directly. Only the signs need care, and they depend on which way you call positive.

In both, $y$ is the coordinate measured from the starting point (y = 0 at t = 0), $u$ is the initial velocity and $v$ the velocity at time t, each with the sign set by your chosen convention.

Motion in two dimensions looks complicated, but it hides a simplification: what happens along x has nothing to do with what happens along y.

Coordinates in a plane#

When a particle is free to move in a plane, call it the x–y plane and pick any two perpendicular directions in it as the axes. Start the clock at a convenient moment and put the origin where the particle is at t = 0. Its position at any time is then fully specified by the pair $(x, y)$.

Between $t$ and $t + \Delta t$ the particle moves from $(x, y)$ to $(x + \Delta x, y + \Delta y)$, so its displacement splits into two perpendicular parts:

Differentiating once more gives the acceleration, component by component:

The big idea: split it in two#

The relations between $x$, $v_x$, $a_x$ are exactly the straight-line relations from earlier. The same is true for $y$, $v_y$, $a_y$. So if the acceleration is constant, each axis obeys its own set of equations:

A cricket ball, a water fountain, a long jumper: anything launched at an angle traces the same family of curves. They are all one horizontal motion and one vertical motion, running side by side.

What makes a projectile#

Throw something at an angle near the Earth’s surface and it follows a curved path. Such an object is a projectile. We will assume it stays close to the ground and that air resistance is negligible. Then its acceleration is constant: $g$, straight down. So projectile motion is just motion in a plane with constant acceleration, exactly what the previous lesson handled.

Splitting the motion#

Put the origin at the launch point, the x-axis horizontal and the y-axis vertically up, and start the clock at launch. With launch speed $u$ at angle $\theta$:

Horizontal: steady

With no horizontal acceleration, the horizontal velocity never changes:

Vertical: thrown straight up

Vertically it is exactly a ball thrown straight up with speed $u\sin\theta$:

Three results worth knowing#

Time of flight

It lands when $y = 0$ again: $0 = t\,(u\sin\theta - \tfrac12gt)$. One root, $t = 0$, is the launch. The other is the landing:

Range

Horizontal distance covered in that time, using $2\sin\theta\cos\theta = \sin 2\theta$:

Maximum height

At the top the velocity is momentarily horizontal, so $v_y = 0$, which happens at $t = u\sin\theta/g$ (3.24), exactly half the time of flight. The climb and the fall take equal times. Substituting:

Position, velocity and acceleration all depend on who is measuring. This lesson gives the rules for translating between observers, and explains why you tilt your umbrella when you run through rain.

Positions in two frames#

Let frames S and S′ watch the same particle P. The frames may be moving relative to each other. With O and O′ their origins, the position of P relative to S is $\vec r_{P,S} = \vec{OP}$ and relative to S′ is $\vec r_{P,S'} = \vec{O'P}$, while the position of S′ itself (its origin) relative to S is $\vec r_{S',S} = \vec{OO'}$. The triangle rule links them:

Velocities in two frames#

Differentiate (3.26) with respect to time:

The velocity of the particle relative to S is its velocity relative to S′ plus the velocity of S′ relative to S. Rearranged, it answers the everyday question “how fast is A moving as seen from B?”:

Two everyday cases. A rifle fires bullets at 80 m/s relative to the rifle. Mounted on a train moving at 25 m/s and fired forward, the bullet moves at 80 + 25 = 105 m/s relative to the ground. And a swimmer who manages 1.5 m/s in still water (that is, relative to the water) swimming with a 1.0 m/s current moves at 2.5 m/s relative to the bank.

Accelerations in two frames#

Differentiate once more:

If S′ moves relative to S at constant velocity, then $\vec a_{S',S} = 0$ and $\vec a_{P,S} = \vec a_{P,S'}$. Observers drifting past each other uniformly disagree about positions and velocities, but they agree about accelerations. That fact will become the foundation of Newton’s laws in the chapters ahead.