Suggest a suitable physical situation for each of the following graphs (Fig 2.12): ![Figure 2.12](/figures/motion-in-a-straight-line/three-graphs-position-velocity-acceleration-suitable-situation-fig-2-12.svg)

Suggest a suitable physical situation for each of the following graphs (Fig 2.12):

Exercises · 2.15

Q. Suggest a suitable physical situation for each of the following graphs (Fig 2.12):

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Concept understanding — One Dimensional Motion

One Dimensional Motion

Imagine you're standing on a long, perfectly straight railway track. A train moves along it — it can only go forward or backward. It cannot turn left, right, up, or down. That's the core idea: motion confined to a single straight line.

The Intuition

In the real world, a ball thrown across a room moves in three dimensions — it goes forward, sideways, and up-down. But many problems in physics are simpler. We deliberately restrict motion to one dimension (1D) to understand the fundamental laws without the clutter of angles and curves.

Think of:

A car moving on a straight highway (no turns).
A lift going up or down a shaft.
A ball dropped straight down from a height.
A puck sliding on a frictionless straight track.

In each case, the object's position can be described by just one number — its distance from a fixed point (the origin) along that line.

The Precise Statement

x

This means:

The path is a straight line.
The direction is either positive (say, to the right or upward) or negative (left or downward).
All vector quantities (displacement, velocity, acceleration) have only two possible directions — forward or backward.

The Three Key Quantities

To describe 1D motion precisely, we use three quantities:

Position ( $x$ or $y$ ) — where the object is relative to the origin.

Example: $x = +5\ \text{m}$ means 5 metres to the right of the origin.
Displacement ( $\Delta x$ ) — change in position:

\Delta x = x_{\text{final}} - x_{\text{initial}}

This is a vector — it has a sign. If you move from $x=2\ \text{m}$ to $x=7\ \text{m}$ , $\Delta x = +5\ \text{m}$ . If you move back to $x=3\ \text{m}$ , $\Delta x = -4\ \text{m}$ .

Velocity ( $v$ ) — rate of change of position:

v = \frac{\Delta x}{\Delta t}

Average velocity has a sign. Instantaneous velocity is the slope of the position-time graph.

Acceleration ( $a$ ) — rate of change of velocity:

a = \frac{\Delta v}{\Delta t}

Again, a signed quantity. Positive acceleration doesn't always mean speeding up — it means velocity is becoming more positive (or less negative).

The Equations of Motion (Constant Acceleration)

For the special (and very common) case of constant acceleration, we have three equations that connect these quantities. They are the equations of motion for 1D:

v = u + at

Where:

$u$ = initial velocity
$v$ = final velocity
$a$ = constant acceleration
$t$ = time
$s$ = displacement

Watch out

These equations only work when acceleration is constant. If acceleration changes, you cannot use them directly — you'd need calculus or graphical methods.

A Simple Example …

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Each graph is the fingerprint of a common motion. Read off what position, velocity or acceleration does with time and name a matching situation: (a) an object thrown up and coming down past its start, (b) a ball bouncing with decreasing speed, and (c) a uniformly moving object hit briefly.

Concept

The three graphs describe the same kind of quantity trio — position, velocity, acceleration — versus time. To 'suggest a physical situation', identify the sign and rate of change in each phase and match it to a real motion whose $x$ , $v$ or $a$ behaves the same way.

(a) Position-time: rise to A, fall to B (below the start)

$x$ is zero, then rises (moving in $+x$ ), reaches a maximum at A (momentarily at the top), then falls, crosses the start level and goes negative to B, then stays constant (at rest).

This matches an object thrown vertically upward from a raised platform (taking upward as $+x$ with the launch point as origin): it rises to its highest point A, then falls back, passing the launch level and continuing down to the ground at a lower position B, where it comes to rest. (The straight segments are the idealised, uniform-velocity version of such up-and-down motion.)

(b) Velocity-time: reversing saw-tooth with shrinking magnitude

$v$ changes linearly (constant slope $=$ constant acceleration) during each segment, and at the end of each segment it suddenly reverses sign while its magnitude drops. A sudden velocity reversal is an impact; a constant slope between impacts is free flight under gravity; shrinking magnitude means energy lost per impact.

This matches a ball released and bouncing repeatedly on the floor: between bounces gravity changes its velocity uniformly, at each strike with the ground the velocity reverses direction, and because the collisions are not perfectly elastic the speed after each bounce is smaller.

(c) Acceleration-time: single short positive pulse …

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Concept: Deriving Each Scenario by Integrating the Given Curve

Method: Work Forward Through the Calculus Relationships (derive what a shape implies, rather than matching it to a remembered scenario)

Instead of recognising each graph shape as "looking like" a familiar motion, this method works forward mathematically — using the fact that $v=dx/dt$ and $a=dv/dt$ — to derive exactly what kind of physical event each shape requires, and only then names a scenario that satisfies that requirement.

(c) Acceleration-time: a single narrow spike

Integrate the spike once to get $v(t)$ : $\displaystyle \Delta v = \int a\,dt$ . Because the spike is narrow (short duration) but has some finite height, this integral is a finite, non-zero number even though the time interval is tiny — this is the signature of an impulsive force. Outside the spike, $a=0$ , so $v(t)$ is exactly constant before the spike and exactly constant (at a new value) after it: a sudden jump in $v$ .
Integrate again to get $x(t)$ : a constant-then-jumped-constant velocity produces an $x$ - $t$ graph that is a straight line of one slope, then (at the spike) a straight line of a different slope — a kink, but no discontinuity in $x$ itself (position can't jump instantaneously).
Conclusion from the calculus alone: this is exactly the fingerprint of a body moving at constant velocity that receives a very brief, strong push — a uniformly-moving object struck for a short time (e.g. a ball struck sharply by a bat while already in flight, or a car briefly tapping its brakes/accelerator).

(b) Velocity-time: reversing sawtooth, shrinking magnitude

Each straight segment has one constant slope (constant $a$ ) — differentiating a straight $v(t)$ segment gives the same constant $|a|=g$ on every segment (each free-fall drop between bounces experiences the same gravity). At the end of each segment, $v$ reverses sign discontinuously while shrinking in magnitude — a discontinuous velocity jump (not position) is exactly what an inelastic collision with a fixed surface produces, with the post-bounce speed reduced (energy lost) each time.
Conclusion: a ball dropped and bouncing repeatedly on the floor, losing some speed at each bounce.

(a) Position-time: rise to a sharp peak A, fall through the start level to B, then constant …

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Q.Suggest a suitable physical situation for each of the following graphs (Fig 2.12):

Concept

(a) Position-time: rise to A, fall to B (below the start)

(b) Velocity-time: reversing saw-tooth with shrinking magnitude

(c) Acceleration-time: single short positive pulse …

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