Motion in One Dimension

What is Kinematics?

Kinematics describes the motion of objects. It analyzes position, velocity, and acceleration without considering the forces that cause the motion. To describe motion precisely, we need to first establish some fundamental concepts.

Observer and Frame of Reference

Consider a reader sitting in a room. The reader appears to be at rest. But the room is on the Earth, and the Earth revolves around the Sun, which in turn moves through the galaxy. Whether a body is at rest therefore depends on who observes it.

This reveals a profound truth: motion is not an absolute property of the object itself. Rather, it is a combined property of the object and the observer. A state of motion cannot be defined without specifying who is observing it. To describe motion, we define an observer. The observer measures the particle's position, velocity, and acceleration.

Associated with an observer is a frame of reference. A frame of reference is a coordinate system (such as $x$, $y$, and $z$ axes) combined with a clock to measure time. The coordinate system specifies the particle's position. The clock records the corresponding time of each measurement.

Rest and Motion: Relative Concepts

Because motion depends on the observer, the concepts of rest and motion are relative. An object is at rest if its position does not change with respect to a chosen frame of reference over time. Conversely, it is in motion if its position does change with respect to that frame of reference.

There is no such thing as absolute rest or absolute motion. For example, a passenger in a moving train is at rest relative to the carriage. However, the passenger is in motion relative to the ground outside. Both descriptions are equally valid, depending on the chosen frame of reference.

The Point Particle

To simplify our analysis of motion, we often use the idealized concept of a point particle. A point particle is an object considered to have mass but no size, shape, or internal structure. This allows us to ignore complex details like rotation or internal vibrations and focus solely on its translational motion (moving from one place to another).

While real objects always have size, the point particle model is an excellent approximation when the object's dimensions are negligible compared to the distances traveled. For example, Earth is treated as a point particle when calculating its orbit around the Sun. The orbital distance is much larger than Earth's radius.

Key Takeaways

• Kinematics describes how objects move without considering why they move.
• Rest and motion are relative concepts, depending on the chosen frame of reference.
• A point particle is an idealized object with mass but no size, useful when an object's dimensions are small compared to the distances involved in its motion.
• An observer measures motion from a specific frame of reference. This includes a coordinate system and a clock.
• Different frames of reference can give different but equally valid descriptions of the same motion.

Distance and Speed

Imagine you want to travel from point A to point B. If there is a winding, curved road connecting them, the actual path length you cover along that road is the distance you travel. Distance is a measure of the actual path length traversed by an object between two points during its motion. It is a scalar quantity, meaning it has only magnitude and no direction. It depends entirely on the path taken, not just the starting and ending points.

The extreme case: a round trip

Push the two measures as far apart as they will go. Send the runner out from A to the turning point B and then straight back to A, the same 100 m of road, covered twice.

Average Speed for Variations in Speed

Case 1: Motion in Equal Time Intervals

Suppose a car travels at speed $v_1$ for a time interval $t$. Then at speed $v_2$ for another equal time interval $t$. What is its average speed?

First, we find the distance covered in each interval: $d_1 = v_1 t$ and $d_2 = v_2 t$. The total distance is $d_1 + d_2 = (v_1 + v_2)t$, and the total time is $2t$. Substituting these into the definition of average speed:

$$\qquad v_{\text{avg}} = \dfrac{\text{Total Distance}}{\text{Total Time}} = \dfrac{(v_1 + v_2)t}{2t} = \dfrac{v_1 + v_2}{2}$$

Thus, when the time intervals are equal, the average speed is simply the arithmetic mean of the individual speeds.

Example: A car travels at $40 \, \text{km/h}$ for $1 \, \text{hour}$ and then at $60 \, \text{km/h}$ for the next $1 \, \text{hour}$. The average speed is $\frac{40 + 60}{2} = 50 \, \text{km/h}$.

Case 2: Motion in Unequal Time Intervals

If the time intervals are not equal, say, traveling at speed $v_1$ for time $t_1$ and speed $v_2$ for time $t_2$, the average speed is the weighted average:

$$\qquad v_{\text{avg}} = \dfrac{v_1 t_1 + v_2 t_2}{t_1 + t_2}$$

Average speed is always total distance over total time. The distance contributed by each speed is that speed multiplied by how long it was held. So each speed is weighted by the time spent at it, not by the distance covered, hold a speed twice as long and it counts twice.

Example: A train travels at $70 \, \text{km/h}$ for $2 \, \text{hours}$ and then at $50 \, \text{km/h}$ for $3 \, \text{hours}$. The average speed is:

$$\qquad v_{\text{avg}} = \dfrac{(70 \times 2) + (50 \times 3)}{2 + 3} = \dfrac{140 + 150}{5} = 58 \, \text{km/h}$$

Case 3: Motion for Equal Distances

Suppose a runner covers a distance $d$ with speed $v_1$, and then covers an equal distance $d$ with speed $v_2$. What is the average speed?

Here, the time taken for each part of the journey is different: $t_1 = \frac{d}{v_1}$ and $t_2 = \frac{d}{v_2}$. The total time is $t_1 + t_2$, and the total distance is $2d$. Thus:

$$\qquad v_{\text{avg}} = \dfrac{2d}{\dfrac{d}{v_1} + \dfrac{d}{v_2}} = \dfrac{2d}{d\left(\dfrac{1}{v_1} + \dfrac{1}{v_2}\right)} = \dfrac{2v_1 v_2}{v_1 + v_2}$$

In this case, where the distances are equal, the average speed is the harmonic mean of the individual speeds. Note that the average speed is always closer to the slower speed because the object spends more time traveling at the slower speed.

Example: A person walks $10 \, \text{km}$ at $4 \, \text{km/h}$ and then another $10 \, \text{km}$ at $6 \, \text{km/h}$. The average speed is:

$$\qquad v_{\text{avg}} = \dfrac{2 \times 4 \times 6}{4 + 6} = \dfrac{48}{10} = 4.8 \, \text{km/h}$$

Case 4: Motion for Unequal Distances

If a particle travels distance $d_1$ at speed $v_1$ and distance $d_2$ at speed $v_2$, the average speed is:

$$\qquad v_{\text{avg}} = \dfrac{d_1 + d_2}{\dfrac{d_1}{v_1} + \dfrac{d_2}{v_2}}$$

Example: A plane flies $3000 \, \text{km}$ at $600 \, \text{km/h}$ and then $2000 \, \text{km}$ at $500 \, \text{km/h}$. The average speed is:

$$\qquad v_{\text{avg}} = \dfrac{3000 + 2000}{\dfrac{3000}{600} + \dfrac{2000}{500}} = \dfrac{5000}{5 + 4} \approx 555.56 \, \text{km/h}$$

Why the two averages differ

Displacement

Displacement, unlike distance, is a vector quantity. It measures the change in position of an object, considering both the magnitude and the direction of that change. It depends solely on the initial and final positions of the object, completely ignoring the path taken in between.

Average Velocity

Average velocity is defined as the net displacement covered by a particle in a given time interval divided by the duration of that time interval:

$$\qquad \text{Average Velocity} = \dfrac{\text{Displacement}}{\text{Time Interval}}$$

1D Motion Example

Consider a particle that starts at $x = 0$, moves to $x = +5\, \text{m}$ in 2 seconds. Then turns back to reach $x = +2\, \text{m}$ at the end of another 3 seconds. Let us find its average velocity for the entire 5-second journey:

• Displacement during the first part (0 to 2s) = $+5\, \text{m}$
• Displacement during the second part (2s to 5s) = $+2\, \text{m} - (+5\, \text{m}) = -3\, \text{m}$
• Total net displacement (0 to 5s) = $+2\, \text{m}$
• Average velocity (0 to 5s) = $\dfrac{+2\, \text{m}}{5\, \text{s}} = +0.4\, \text{m/s}$

2D Motion Example

A particle moves in the $xy$-plane. At time $t_1 = 0$, its position vector is $\vec{r}_1 = 2\hat{i} + 3\hat{j}$ m. At time $t_2 = 4\, \text{s}$, its position is $\vec{r}_2 = 10\hat{i} - 3\hat{j}$ m. Let us calculate its average velocity vector:

First, we find the displacement vector $\Delta\vec{r}$:

$$\qquad \Delta\vec{r} = \vec{r}_2 - \vec{r}_1 = (10 - 2)\hat{i} + (-3 - 3)\hat{j} = 8\hat{i} - 6\hat{j} \, \text{m}$$

Next, we divide by the time interval $\Delta t = 4\, \text{s}$ to find the average velocity vector:

$$\qquad \vec{v}_{\text{avg}} = \dfrac{\Delta\vec{r}}{\Delta t} = \dfrac{8\hat{i} - 6\hat{j}}{4} = 2\hat{i} - 1.5\hat{j} \, \text{m/s}$$

The magnitude of this average velocity is:

$$\qquad |\vec{v}_{\text{avg}}| = \sqrt{(2)^2 + (-1.5)^2} = \sqrt{4 + 2.25} = \sqrt{6.25} = 2.5 \, \text{m/s}$$

3D Motion Example

A drone is spotted at $\vec{r}_1 = \hat{i} + \hat{j} + 6\hat{k}\ \text{m}$. Two seconds later it is at $\vec{r}_2 = 7\hat{i} + 4\hat{j} + 4\hat{k}\ \text{m}$. Find its average velocity over that interval.

Nothing is said about the route it took, and nothing needs to be. The displacement is the difference of the two position vectors:

Its magnitude is $|\Delta\vec{r}| = \sqrt{6^2 + 3^2 + (-2)^2} = \sqrt{49} = 7\,\text{m}$, the drone finished $7\,\text{m}$ from where it started, in a straight line.

Dividing by the elapsed time gives the average velocity:

with magnitude $|\vec{v}_{\text{avg}}| = \sqrt{9 + 2.25 + 1} = 3.5\,\text{m/s}$. Notice the minus sign on $\hat{k}$: the drone finished lower than it started. Consequently, the vertical component of its average velocity points downwards.

Whatever route the drone flies between the two points, its displacement is the one straight arrow joining them. Drag the scene to turn it and see the same arrow from any direction.

Drag the picture to turn it. A flat drawing of a three-dimensional situation always leaves the same doubt, is that arrow really straight, or does it only look straight from this one angle? Turning it settles the question: the purple arrow stays a single straight segment from every viewpoint, while the dashed route bends differently as you go round.

The dashed curve is a route the drone might have taken; there are infinitely many others. Every single one of them gives the same $\Delta\vec{r}$. Therefore the same average velocity, because average velocity asks only where you started and where you finished. That is exactly what makes it cheap to compute and, on its own, a poor description of a journey.

Speed vs Velocity

While speed and velocity are related, they are fundamentally different:

• Speed is a scalar (magnitude only) and is based on distance.
• Velocity is a vector (magnitude and direction) and is based on displacement.

This means:

• Average speed can never be negative, as it is calculated from the total distance traveled.
• Average velocity can be positive, negative, or zero, reflecting the direction of displacement.
• For the same motion, the average speed and the magnitude of the average velocity may differ if the path is not a straight line in one direction.

Key Differences Summarized

The Idea of 'Instantaneous'

Instantaneous speed represents how fast an object is moving at a specific instant in time. It is obtained from average speed by shrinking the time interval until it becomes extremely small. This allows us to measure the rate of change of position at an exact moment.

Consider a particle moving according to the equation $x = t^2$, where $x$ is the position in meters and $t$ is the time in seconds. To better understand this motion, the following table illustrates the particle's position at specific times:

From the table, we can see the particle is speeding up. Let us compare average velocities over different intervals, each one is $\Delta x / \Delta t$:

Average velocity from $t = 0$ to $t = 4$:

$$\text{Average velocity} = \dfrac{16 - 0}{4 - 0} = 4 \, \text{m/s}$$

Average velocity from $t = 2$ to $t = 5$:

$$\text{Average velocity} = \dfrac{25 - 4}{5 - 2} = 7 \, \text{m/s}$$

Shrinking the interval

We want the speed at the single instant $t = 2\,\text{s}$. The average-speed formula cannot be asked that question directly, because it needs an interval to divide by. An instant has no duration, the formula would read $0/0$.

So do the next best thing. Take a short interval that starts at $t = 2\,\text{s}$, work out the average speed over it, and then make it shorter. Call the length of that interval $h$, so the interval runs from $t = 2$ to $t = 2 + h$.

The answers are closing in on $4$. Each added row gives a more precise answer. We could continue adding rows, but notice what the table can never do. Every row still has a real, non-zero $h$ in it. Consequently, every row is still an average over a stretch of time, not a speed at an instant. Adding rows gets us closer and never gets us there.

What we need is a way of doing every row at once.

Doing every row at once

Rather than choosing a number for $h$, keep it as a symbol and see what comes out. With $x = t^2$, the average speed from $t = 2$ to $t = 2 + h$ is

$$v_{\text{avg}} = \dfrac{(2+h)^2 - 2^2}{h} = \dfrac{4 + 4h + h^2 - 4}{h} = \dfrac{4h + h^2}{h} = \dfrac{h\,(4 + h)}{h} = 4 + h$$

One line has replaced the whole table. Put $h = 1$ and it gives $5$; put $h = 0.1$ and it gives $4.1$; put $h = 0.01$ and it gives $4.01$. Every row we computed, and every row we did not, is contained in $4 + h$.

Newton's way out: $h$ tending to zero

Newton resolved this difficulty, and the idea he introduced is the one still used today. We do not set $h$ to zero. We let $h$ tend to zero, written $h \to 0$. The phrase is doing two jobs at once:

• $h$ is never zero, so dividing by it, and cancelling it, is legal at every stage of the argument;
• $h$ becomes smaller than any number you can name, so once the cancelling is done, adding $h$ to $4$ can no longer change the answer to anything distinguishable from $4$.

Both halves are needed, and they act in that order: non-zero while we divide, negligible once we have divided. Written out, the speed at $t = 2\,\text{s}$ is

$$v = \lim_{h \to 0} \left( 4 + h \right) = 4\ \text{m/s}$$

and the same argument at a general time $t$ gives $v = \lim_{h\to 0}(2t + h) = 2t$. At $t = 2$ that is $4\,\text{m/s}$, in agreement with the table.

The same limit, drawn

Everything above happens on the position-time graph of $x = t^2$ as well. It is worth seeing, because it turns the limit from a piece of algebra into a picture you can hold on to.

Mark $P$ at $t = 2$. For each $h$, mark $Q$ at $t = 2 + h$ and join them. The straight line $PQ$ is a chord. Its slope, rise over run, is exactly the average speed we tabulated: $4 + h$.

Instantaneous Velocity and the Derivative

This process leads to the formal definition of instantaneous velocity. Mathematically, if $x$ represents the position of an object at time $t$, the instantaneous velocity $v$ is defined as the limit of the average velocity $\Delta x / \Delta t$ as the time interval approaches zero:

$$v = \lim_{\Delta t \to 0} \dfrac{\Delta x}{\Delta t} = \dfrac{dx}{dt}$$

The instantaneous speed is the magnitude of this, $|v|$; for the motion in this example ($x = t^2$ with $t > 0$) the particle never reverses, so the two are the same number here.

This formula introduces the concept of differentiation, represented by $\dfrac{dx}{dt}$. This means "the rate of change of position ($x$) with respect to time ($t$)."

For $x=t^2$, we found that at any time $t$, the instantaneous velocity is $v = 2t$. This is the derivative of $t^2$:

$$\dfrac{d}{dt}(t^2) = 2t$$

Basic Differentiation Formulas

One tangent per instant

There was nothing special about $t = 2$. The same argument runs at any time $t$. It gives a tangent at every point of the curve, so the slope is not one number, it is a function of time. That is what $v = 2t$ is saying.

Application Examples

Instantaneous Velocity

Instantaneous velocity, similar to instantaneous speed, is the velocity at a specific moment. It is the rate of change of displacement with respect to time, calculated using differentiation:

$$v = \dfrac{dx}{dt}$$

The difference is that instantaneous velocity is a vector quantity. Consequently, it includes direction (positive or negative in one dimension), whereas instantaneous speed is the magnitude of the velocity and is always positive.

What is Acceleration?

Acceleration is the rate of change of velocity with respect to time. It is a vector quantity, meaning it has both magnitude and direction.

Average Acceleration

Average acceleration is the change in velocity in a given time interval divided by the time interval over which the change occurs:

$$\qquad a_{\text{avg}} = \dfrac{\Delta v}{\Delta t} = \dfrac{v_f - v_i}{t_f - t_i}$$

where $v_i$ and $v_f$ are the initial and final velocities at times $t_i$ and $t_f$ respectively.

Example 1: If a car's velocity changes from $+20 \, \text{m/s}$ to $+30 \, \text{m/s}$ in $5 \, \text{s}$, then its average acceleration is:

$$\qquad a_{\text{avg}} = \dfrac{30 - 20}{5} = 2 \, \text{m/s}^2$$

Example 2: If the velocity changes from $+10 \, \text{m/s}$ to $-5 \, \text{m/s}$ in 2 seconds, then:

$$\qquad a_{\text{avg}} = \dfrac{-5 - 10}{2} = -7.5 \, \text{m/s}^2$$

The negative sign signifies that the acceleration is in the opposite direction to the initial velocity.

What acceleration actually looks like

Two carts set off together. Both cover $40\,\text{m}$ and both take $5\,\text{s}$, so their average velocities are identical, $8\,\text{m/s}$ each. Everything that distinguishes them is inside the journey.

Instantaneous Acceleration

Instantaneous acceleration is the acceleration at a specific instant in time. It is the limit of the average acceleration as the time interval approaches zero. Just as instantaneous velocity is the derivative of position with respect to time, instantaneous acceleration is the derivative of velocity with respect to time:

$$\qquad a = \lim_{\Delta t \to 0} \dfrac{\Delta v}{\Delta t} = \dfrac{dv}{dt}$$

Since velocity is the derivative of position ($v = dx/dt$), acceleration can also be expressed as the second derivative of position with respect to time:

$$\qquad a = \dfrac{d}{dt}\left(\dfrac{dx}{dt}\right) = \dfrac{d^2x}{dt^2}$$

Example: If the velocity $v(t)$ in m/s of a particle depends on time $t$ in seconds given by $v(t) = 3t^2 - 2t + 1$, its instantaneous acceleration is:

$$\qquad a(t) = \dfrac{dv}{dt} = 6t - 2$$

• At $t=1$ s, $a(1) = 6(1) - 2 = 4 \, \text{m/s}^2$
• At $t=2$ s, $a(2) = 6(2) - 2 = 10 \, \text{m/s}^2$

Understanding Signs of Acceleration

The sign of acceleration tells us about the nature of motion:

• Positive acceleration (+a): Velocity is increasing in the positive direction, or decreasing in the negative direction
• Negative acceleration (-a): Velocity is decreasing in the positive direction, or increasing in the negative direction

Important: Negative acceleration does not always mean "slowing down". It depends on the direction of velocity:

• If $v > 0$ and $a < 0$: Object is slowing down
• If $v < 0$ and $a < 0$: Object is speeding up (in negative direction)
• If $v > 0$ and $a > 0$: Object is speeding up (in positive direction)
• If $v < 0$ and $a > 0$: Object is slowing down

A case where both are negative

More Examples of Acceleration

Uniformly Accelerated Motion

Uniformly accelerated motion means that the acceleration of an object remains constant over time. It does not change in magnitude or direction. In such cases, we can use a set of equations to describe the motion. These are known as the equations of motion.

Important Note: These equations are valid only for uniformly accelerated motion.

First Equation of Motion: $v = u + at$

If acceleration $a$ is constant, the change in velocity $\Delta v$ over a time interval $t$ is simply $a \times t$. If the initial velocity is $u$ at time $t=0$ and the final velocity after time $t$ is $v$, then $\Delta v = v - u$.

$$\qquad v - u = at$$

$$\qquad \Rightarrow v = u + at$$

Second Equation of Motion: $s = ut + \frac{1}{2}at^2$

For uniformly accelerated motion, the average velocity is given by:

$$\qquad v_{\text{avg}} = \frac{u + v}{2}$$

The displacement $s$ is the average velocity multiplied by the time $t$:

$$\qquad s = v_{\text{avg}} \cdot t = \left(\frac{u+v}{2}\right) t$$

Substituting $v = u + at$ from the first equation:

$$\qquad s = \left(\frac{u + (u + at)}{2}\right) t = \left(\frac{2u+at}{2}\right)t$$

$$\qquad \Rightarrow s = ut + \frac{1}{2}at^2$$

Third Equation of Motion: $v^2 = u^2 + 2as$

From the first equation, we can express time as $t = \frac{v - u}{a}$. Substituting this into the equation $s = \left(\frac{u+v}{2}\right) t$:

$$\qquad s = \left(\frac{u+v}{2}\right) \left(\frac{v-u}{a}\right) = \frac{v^2 - u^2}{2a}$$

$$\qquad \Rightarrow 2as = v^2 - u^2$$

$$\qquad \Rightarrow v^2 = u^2 + 2as$$

Summary of Equations for Uniformly Accelerated Motion

Where these equations stop working

Press let h → 0 on the figure above and watch the secant roll onto the tangent at $P$. It never arrives at some particular small interval, it approaches the tangent. The average velocity approaches the instantaneous one at the same rate.

Thus, we conclude that:

The slope of a tangent drawn at a point in position time graph, represents the instantaneous velocity of associated particle at that time.

This relation of slope can be extended to any graph in general. For example, the slope of a tangent drawn at a point in velocity time graph, represents the instantaneous acceleration of associated particle at that time. In general,

The slope of a tangent drawn at a point in any general graph, represents the rate of change of quantity on $y$-axis with respect to $x$-axis.

Let us consider a velocity-time graph where a particle's velocity changes in steps:

Press the buttons on the figure above and read the running total; the arithmetic underneath is the same sum written out.

The total displacement becomes $24 \,\text{m}$. Notice that each term in the sum is the area of a rectangular section under the velocity-time graph. This leads to a general principle:

The total displacement is the area under the velocity-time graph

If the velocity changes continuously (not in steps), we can still apply this principle. We divide the time into very small intervals $\Delta t$. Over each small interval, we assume the velocity remains essentially constant. The displacement during each interval is approximately $v \Delta t$ (the area of a very thin rectangle). Summing up these small displacements (or areas) gives an approximation of the total displacement.

As $\Delta t$ becomes infinitesimally small, this sum approaches the exact area under the curve, which is the precise displacement. This is shown in animated graph below:

Real velocities do not change in steps, so cover the area with rectangles instead and make them narrower. Each button halves their width:

Two things about that pair of moves are worth committing to memory, because they are what most graph questions are quietly testing. Going up the chain by taking an area gives you the change in a quantity, not the quantity itself, you still need a starting value before an area becomes an answer. And area below the axis counts as negative, which is precisely how a displacement can shrink while the distance travelled keeps growing.