Motion in a Straight Line • Topic 1 of 3

Position, Distance & Displacement

Mechanics begins with kinematics — the description of motion without asking what causes it. The simplest case is motion along a single straight line, called rectilinear motion or motion in one dimension. To describe even this we must first agree on a way to say where an object is, which is why every problem starts with a frame of reference.

A frame of reference is a coordinate system attached to an observer, together with a clock. For 1D motion we draw a single axis (usually the $x$-axis), pick an origin $O$ as the zero mark, and choose a positive direction. The position of a particle is then a single number $x$ — positive on one side of the origin, negative on the other. A car standing at $x = +200\,\text{m}$ and another at $x = -150\,\text{m}$ are on opposite sides of $O$. The same motion looks different in different frames: a ball dropped inside a moving train falls straight down for a passenger but follows a curved path for someone on the platform.

As a particle moves, two quantities describe how far it has gone. The path length (or distance) is the total length of the actual path travelled. It is a scalar, is always positive, and never decreases. The displacement $\Delta x = x_2 - x_1$ is the change in position — the straight gap from start to finish with a sign that tells the direction. In one dimension a vector is fully captured by this sign: $+$ means one way, $-$ the other. So displacement is a vector in 1D, while path length is a scalar.

The two are equal in size only when motion is in a single direction without reversing. The moment a particle turns back, path length keeps growing while the magnitude of displacement may shrink. A boy who walks from $x = 0$ to $x = 60\,\text{m}$ and back to $x = 20\,\text{m}$ covers a path length of $60 + 40 = 100\,\text{m}$, but his displacement is only $+20\,\text{m}$. Thus path length $\ge |\Delta x|$ always.

Motion is called uniform when the particle covers equal displacements in equal intervals of time, however small the interval. Otherwise it is non-uniform. A metro coasting on a clear track is close to uniform; the same metro pulling out of a station is non-uniform.

Frame of reference: origin + axis + clock; chosen before describing any motion.
Position $x$: a signed number; sign shows side of origin.
Path length: total path travelled; scalar; always increases.
Displacement $\Delta x = x_2 - x_1$: vector in 1D; can be $+$, $-$ or $0$.
Always $|\Delta x| \le$ path length; equal only if direction never reverses.

Solved Examples

Worked Example

A particle moves along the $x$-axis from $x_1 = -4\,\text{m}$ to $x_2 = +12\,\text{m}$. Find its displacement.

Solution

Displacement $\Delta x = x_2 - x_1$.
$\Delta x = 12 - (-4) = 12 + 4 = 16\,\text{m}$.
The sign is positive, so the shift is along the $+x$ direction.

Answer: $\Delta x = +16\,\text{m}$ (along $+x$).

Worked Example

A boy walks from $x = 0$ to $x = 70\,\text{m}$ and then returns to $x = 25\,\text{m}$. Find the path length and the displacement.

Solution

Forward path $= 70\,\text{m}$; return path $= 70 - 25 = 45\,\text{m}$.
Path length $= 70 + 45 = 115\,\text{m}$.
Displacement $= x_{final} - x_{initial} = 25 - 0 = 25\,\text{m}$.

Answer: Path length $= 115\,\text{m}$, Displacement $= +25\,\text{m}$.

Worked Example

A runner completes one full lap of a $400\,\text{m}$ straight-and-curved track and stops at the start. State the path length and the displacement.

Solution

One full lap means the actual path travelled $= 400\,\text{m}$.
The start and end points coincide.
Displacement $= $ shortest gap between identical points $= 0$.

Answer: Path length $= 400\,\text{m}$, Displacement $= 0$.

Worked Example

Two cars are at positions $x_A = +320\,\text{m}$ and $x_B = -180\,\text{m}$ on a straight road. How far apart are they?

Solution

Separation $= |x_A - x_B|$.
$= |320 - (-180)| = |320 + 180|$.
$= 500\,\text{m}$.

Answer: The cars are $500\,\text{m}$ apart.

Worked Example

In successive equal intervals of $2\,\text{s}$, a body covers displacements $6\,\text{m}, 6\,\text{m}, 6\,\text{m}$ along $+x$. Is the motion uniform? Find the total displacement.

Solution

Equal displacements ($6\,\text{m}$) occur in equal intervals ($2\,\text{s}$).
By definition, this is uniform motion.
Total displacement $= 6 + 6 + 6 = 18\,\text{m}$.

Answer: Uniform motion; total displacement $= 18\,\text{m}$.

Worked Example

A particle starts at $x = 5\,\text{m}$, moves to $x = -3\,\text{m}$, then to $x = 9\,\text{m}$. Find the total path length and the net displacement.

Solution

First leg: $|{-3} - 5| = 8\,\text{m}$. Second leg: $|9 - (-3)| = 12\,\text{m}$.
Path length $= 8 + 12 = 20\,\text{m}$.
Net displacement $= x_{final} - x_{initial} = 9 - 5 = 4\,\text{m}$.

Answer: Path length $= 20\,\text{m}$, Displacement $= +4\,\text{m}$.

Key Points

A frame of reference (origin + axis + clock) must be fixed before describing any motion.
Position $x$ is a signed number; its sign tells which side of the origin the particle is on.
Path length is the total path travelled — a scalar that never decreases.
Displacement $\Delta x = x_2 - x_1$ is a vector in 1D and can be positive, negative or zero.
Path length $\ge |\Delta x|$; they are equal only when the direction of motion never reverses.

Quick Check — 4 MCQs

Tap an option to check your answer0 / 4

Q1.A particle moves from $x_1 = -2\,\text{m}$ to $x_2 = 6\,\text{m}$. Its displacement is:

Explanation: $\Delta x = x_2 - x_1 = 6 - (-2) = 8\,\text{m}$, directed along $+x$.

Q2.Which statement is always true for motion in a straight line?

Explanation: Path length is the total path travelled and can never be less than the size of the straight-line displacement.

Q3.An object returns to its starting point after moving around. Over the whole trip, its displacement is:

Explanation: Start and end coincide, so the change in position is zero, even though path length is not.

Q4.Equal displacements in equal time intervals (however small) define:

Explanation: This is precisely the definition of uniform motion in a straight line.

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