Motion and Graphs

Displacement, Velocity, and Acceleration

Displacement is a vector representing the straight-line change in position from initial to final point, while distance is the total path length traveled (a scalar).

$$\vec{d} = \vec{r_2} - \vec{r_1}$$

Displacement is the vector difference between final and initial position vectors.

$\vec{d}$=Displacement vector(m)

$\vec{r_1}$=Initial position vector(m)

$\vec{r_2}$=Final position vector(m)

Object returns to start

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Displacement = 0 even if distance traveled is large

Straight-line motion

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Displacement magnitude equals distance (path coincides with displacement vector)

Key Distinction: A car driving 5 km north then 5 km south has distance = 10 km but displacement = 0 km.

Scalar vs Vector: Distance is always positive; displacement can be negative (indicating direction).

Average velocity is total displacement divided by total time, while instantaneous velocity is the velocity at a specific instant — the limiting value of $\Delta d / \Delta t$ as $\Delta t \to 0$.

$$v_{avg} = \frac{\Delta d}{\Delta t}, \quad v_{inst} = \lim_{\Delta t \to 0} \frac{\Delta d}{\Delta t}$$

Average velocity gives an overall rate of position change; instantaneous velocity gives the rate at a single moment.

$v_{avg}$=Average velocity(m/s)

$v_{inst}$=Instantaneous velocity(m/s)

$\Delta d$=Change in displacement(m)

Uniform velocity

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$v_{avg} = v_{inst}$ at all times

Object returns to start

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$v_{avg} = 0$ regardless of how fast it moved

Graphical Meaning: On an s-t graph, average velocity = chord slope; instantaneous velocity = tangent slope.

Speed vs Velocity: Speed is the magnitude of velocity (always positive). Average speed = total distance / total time, which differs from average velocity when direction changes.

Limiting Case: As $\Delta t \to 0$, average velocity converges to instantaneous velocity. For uniform motion, they are always identical.

Acceleration is the time rate of change of velocity. It occurs whenever there is a change in speed, direction, or both.

$$a = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{\Delta t}$$

Acceleration is how quickly velocity changes — it is a vector in the direction of the velocity change.

$a$=Acceleration(m/s²)

$\Delta v$=Change in velocity(m/s)

$\Delta t$=Time interval(s)

$a > 0$ and $v > 0$

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Object speeds up in positive direction

$a < 0$ and $v > 0$

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Object slows down (deceleration)

$a$ and $v$ have same sign

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Speed increases; opposite signs means speed decreases

Uniform Acceleration: When velocity changes by equal amounts in equal time intervals — average acceleration equals instantaneous acceleration.

Dimensional Check: Acceleration has dimensions $[LT^{-2}]$. If your answer has units like m/s or m, you've made an error.

Interpreting Displacement-Time (s-t) Graphs

A displacement-time graph tracks an object's position relative to an origin; its gradient reveals the object's instantaneous velocity.

$$v = \frac{ds}{dt} \approx \frac{\Delta s}{\Delta t}$$

The slope of the tangent at any point on an s-t graph equals the velocity at that instant.

$v$=Velocity(m/s)

$\Delta s$=Change in displacement(m)

$\Delta t$=Time interval(s)

Graph is horizontal

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Object is stationary ($v = 0$)

Graph is a straight diagonal

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Object moves with constant velocity

Graph is curved

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Object is accelerating or decelerating

Gradient Significance: A steeper slope indicates higher speed; a negative slope indicates motion back toward the origin.

Instantaneous vs Average: The tangent slope gives instantaneous velocity, while the chord between two points gives average velocity.

Curvature: Upward concavity (curve getting steeper) indicates positive acceleration, while downward concavity (curve flattening) indicates deceleration.

Proportionality: For constant acceleration from rest, $s \propto t^2$ — doubling time quadruples displacement.

Velocity-Time (v-t) Graphs: Slope and Area

Velocity-time graphs describe how fast an object moves over time, where the slope represents acceleration and the area under curve represents displacement.

$$a = \frac{dv}{dt}, \quad \Delta s = \int v \, dt$$

The gradient determines the rate of velocity change, while the geometric area equates to the total displacement.

$a$=Acceleration(m/s²)

$\Delta v$=Change in velocity(m/s)

$\Delta s$=Total displacement(m)

Slope is zero (horizontal line)

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Uniform motion (constant velocity, zero acceleration)

Area is below time-axis

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Negative displacement (moving backwards)

Line crosses time-axis

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Object reverses direction at that instant

Acceleration vs Deceleration: A positive gradient implies speeding up in the positive direction; a negative gradient implies deceleration.

Total Distance vs Displacement: Total distance is the sum of absolute areas (all positive), while displacement is the algebraic sum of areas (positive minus negative).

Area Shapes: Triangle = $\frac{1}{2}bh$, Rectangle = $bh$, Trapezium = $\frac{1}{2}(a+b)h$ — these geometric formulas are essential for v-t graph problems.

Proportionality: $a \propto$ slope — doubling acceleration doubles the gradient of the v-t line.

Acceleration-Time (a-t) Graphs

Acceleration-time graphs show the rate of change of velocity; the area under this graph represents the change in velocity.

$$\Delta v = \text{Area under } a\text{-}t \text{ graph}$$

Integrating acceleration over a time period yields the total velocity gained or lost.

$\Delta v$=Change in velocity(m/s)

$a$=Acceleration(m/s²)

$a = 0$

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Velocity remains constant

Constant positive $a$

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v-t graph is a straight line with positive slope; s-t graph is a parabola

Constant Acceleration: Represented by a horizontal line on the a-t graph. This results in a linear v-t graph and a parabolic s-t graph.

Graph Chain: These three graph types are linked by differentiation (slope) and integration (area): s-t → (slope) → v-t → (slope) → a-t, and a-t → (area) → v-t → (area) → s-t.

Limiting Case: For free fall, the a-t graph is a horizontal line at $a = g = 9.8$ m/s² (or $-9.8$ m/s² if upward is positive).

Equations of Uniformly Accelerated Motion

For motion with constant acceleration along a straight line, four kinematic equations relate displacement, velocity, acceleration, and time.

First Equation: $v_f = v_i + at$ — gives final velocity from initial velocity and acceleration.

Second Equation: $s = \frac{v_i + v_f}{2} \times t$ — displacement from average of initial and final velocities.

Third Equation: $s = v_i t + \frac{1}{2}at^2$ — displacement without knowing final velocity.

Fourth Equation: $v_f^2 = v_i^2 + 2as$ — final velocity without knowing time.

Choosing the Right Equation: Identify which variable is missing from the problem, then use the equation that does not contain it.

Equation Selection Guide

Pick the equation based on the unknown you do NOT need

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Missing $s$: use $v_f = v_i + at$

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Missing $a$: use $s = \frac{v_i + v_f}{2} \times t$

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Missing $v_f$: use $s = v_i t + \frac{1}{2}at^2$

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Missing $t$: use $v_f^2 = v_i^2 + 2as$

The distance covered in the $n^{th}$ second of motion (not the first $n$ seconds) is given by a special formula derived from the third equation of motion.

$$s_n = v_i + \frac{a}{2}(2n - 1)$$

This gives the displacement in a single specific second, not the cumulative displacement.

$s_n$=Distance in the nth second(m)

$v_i$=Initial velocity(m/s)

$a$=Constant acceleration(m/s²)

$n$=The specific second (1st, 2nd, 3rd...)(dimensionless)

$v_i = 0$ (starting from rest)

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Distances in successive seconds follow the odd number ratio 1 : 3 : 5 : 7...

$a = 0$ (uniform velocity)

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Distance in every second is the same: $s_n = v_i$

Derivation Sketch: $s_n = s(n) - s(n-1) = [v_i n + \frac{1}{2}an^2] - [v_i(n-1) + \frac{1}{2}a(n-1)^2]$, which simplifies to the formula above.

Proportionality: For $v_i = 0$, the distance in the $n^{th}$ second $\propto (2n - 1)$. This is linear in $n$, not quadratic.

In the absence of air resistance, all objects near the Earth's surface fall with a uniform acceleration $g \approx 9.8$ m/s², regardless of their mass.

Sign Convention: Choose upward or downward as positive and stay consistent. If upward is positive, then $a = -g = -9.8$ m/s².

Symmetry: Time going up = time coming down (same height). Speed at launch = speed on return (same height).

Proportionality: For an object dropped from rest, $s \propto t^2$ — falling 4× the distance in 2× the time.