Kinematics and Graphical Analysis

🟢 Lite — Quick Review (1h–1d)

Rapid summary of kinematics and graphical analysis for NABTEB physics.

Kinematics is the study of motion without considering the forces that cause it. This topic covers motion in a straight line.

Key Quantities:

Quantity	Symbol	Unit	Type
Distance	$s$	metre (m)	Scalar
Displacement	$s$	metre (m)	Vector
Speed	$v$	m/s	Scalar
Velocity	$v$	m/s	Vector
Acceleration	$a$	m/s²	Vector
Time	$t$	second (s)	Scalar

Equations of Motion (for uniform acceleration):

1. $v = u + at$
2. $s = ut + \frac{1}{2}at^2$
3. $v^2 = u^2 + 2as$
4. $s = \frac{(u + v)}{2}t$

Where: $u$ = initial velocity, $v$ = final velocity, $a$ = acceleration, $s$ = displacement, $t$ = time.

Key Concepts:

• Uniform velocity: Equal displacements in equal time intervals
• Uniform acceleration: Equal changes in velocity in equal time intervals
• Free fall: Motion under gravity alone; $a = g = 9.8 , \text{m/s}^2$ (downward)
• Projectile motion: Motion under gravity with initial horizontal velocity

Graphs of Motion:

Displacement-Time Graph:

• Gradient = velocity
• Straight line = uniform velocity
• Curved line = changing velocity

Velocity-Time Graph:

• Gradient = acceleration
• Area under graph = displacement
• Horizontal line = uniform velocity

Acceleration-Time Graph:

• Area under graph = change in velocity

⚡ NABTEB Exam Tip: In velocity-time graph questions, ALWAYS find displacement by calculating the AREA under the graph — even for curved graphs. Break the area into triangles and rectangles for easier calculation.

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🟡 Standard — Regular Study (2d–2mo)

For NABTEB students who want thorough understanding of kinematics and graphical analysis.

Understanding the Equations of Motion:

Derivation of $v = u + at$: By definition of uniform acceleration: $$a = \frac{v - u}{t}$$ Rearranging: $v = u + at$

Derivation of $s = ut + \frac{1}{2}at^2$: Average velocity $= \frac{u + v}{2}$ (for uniform acceleration) Displacement $s = \text{average velocity} \times \text{time} = \frac{u + v}{2}t$ Substituting $v = u + at$: $$s = \frac{u + (u + at)}{2}t = \frac{2u + at}{2}t = ut + \frac{1}{2}at^2$$

Derivation of $v^2 = u^2 + 2as$: From $s = \frac{(u + v)}{2}t$ and $t = \frac{v - u}{a}$: $$s = \frac{(u + v)}{2} \times \frac{(v - u)}{a} = \frac{v^2 - u^2}{2a}$$ Rearranging: $v^2 = u^2 + 2as$

Sign Convention:

• Take the direction of motion as positive
• If acceleration is in the same direction as velocity, $a$ is positive
• If acceleration opposes motion (deceleration), $a$ is negative

Free Fall: When an object is dropped: $u = 0$, $a = +g = 9.8 , \text{m/s}^2$ (downward) When an object is thrown upward: $u$ is positive (upward), $a = -g = -9.8 , \text{m/s}^2$ (downward)

Time to maximum height (for object thrown upward with velocity $u$): $$t = \frac{u}{g}$$

Maximum height: $$h = \frac{u^2}{2g}$$

Projectile Motion:

An object thrown horizontally from height $h$ with speed $u$:

• Horizontal motion: constant velocity $u_x = u$ (no acceleration in horizontal direction)
• Vertical motion: starts at $u_y = 0$, accelerates at $g = 9.8 , \text{m/s}^2$ downward

Time of flight (for horizontal projectile from height $h$): $$t = \sqrt{\frac{2h}{g}}$$

Range (horizontal distance): $$R = u \times t = u\sqrt{\frac{2h}{g}}$$

Graphical Analysis — Detailed:

Displacement-Time Graph:

• Slope at any point = instantaneous velocity
• Positive slope = moving in positive direction
• Negative slope = moving in negative direction
• Zero slope = at rest
• Curving upward = accelerating
• Curving downward = decelerating

Velocity-Time Graph:

• Slope at any point = instantaneous acceleration
• Area above time axis = positive displacement
• Area below time axis = negative displacement
• Zero slope = zero acceleration (constant velocity)

Acceleration-Time Graph:

• Zero everywhere = constant velocity
• Positive value = increasing velocity in positive direction
• Negative value = decreasing velocity (or increasing in negative direction)

Motion in a Circle:

For an object moving in a circle with constant speed:

• Velocity constantly changes direction (so it IS accelerating)
• Centripetal acceleration: $a_c = \frac{v^2}{r} = \omega^2 r$
• Centripetal force: $F_c = \frac{mv^2}{r} = m\omega^2 r$

⚡ NABTEB Exam Tip: When solving projectile motion problems, always treat horizontal and vertical components separately. Horizontal motion has constant velocity; vertical motion has constant acceleration $g = 9.8 , \text{m/s}^2$.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage of kinematics for NABTEB students preparing for exams.

Derivation Using Calculus:

With calculus, kinematics becomes more elegant:

Velocity is the rate of change of displacement: $$v = \frac{ds}{dt}$$

Acceleration is the rate of change of velocity: $$a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$$

From these definitions, we can derive all equations:

Starting with $a = \frac{dv}{dt}$: $$dv = a , dt$$ Integrating: $\int_{u}^{v} dv = a \int_{0}^{t} dt$ $\Rightarrow v - u = at$ $\Rightarrow v = u + at$

Starting with $v = \frac{ds}{dt}$: $$ds = v , dt = (u + at) , dt$$ Integrating: $\int_{0}^{s} ds = \int_{0}^{t} (u + at) , dt$ $\Rightarrow s = ut + \frac{1}{2}at^2$

Relative Motion:

If two objects A and B move with velocities $v_A$ and $v_B$ (in the same direction):

• Relative velocity of A with respect to B: $v_{AB} = v_A - v_B$
• Relative velocity of B with respect to A: $v_{BA} = v_B - v_A$

Non-Horizontal Projectile Motion:

For a projectile launched at angle $\theta$ to the horizontal with speed $u$:

• Horizontal component: $u_x = u \cos\theta$
• Vertical component: $u_y = u \sin\theta$

Time of flight: $$T = \frac{2u \sin\theta}{g}$$

Maximum height: $$H = \frac{u^2 \sin^2\theta}{2g}$$

Range (horizontal distance): $$R = \frac{u^2 \sin 2\theta}{g}$$

Key Observations:

• Range is maximum when $\theta = 45°$ (since $\sin 2\theta = 1$ at $2\theta = 90°$)
• Complementary angles give the same range (e.g., $30°$ and $60°$)
• Time of flight depends only on vertical component and $g$
• Maximum height depends only on vertical component and $g$

Condition for Circular Motion:

An object moving in a circle of radius $r$ with speed $v$ requires a centripetal force: $$F_c = \frac{mv^2}{r}$$

If this force is removed, the object flies off tangentially (tangent to the circle at the point of release) — Newton’s First Law.

Vertical Circular Motion:

For an object in vertical circular motion (like a roller coaster):

• At the bottom: $T - mg = \frac{mv^2}{r}$ (tension must exceed weight)
• At the top: $T + mg = \frac{mv^2}{r}$ (tension adds to weight)
• Minimum speed at top to complete circle: $v_{\min} = \sqrt{gr}$
• Minimum speed at bottom to complete loop: $v_{\min} = \sqrt{5gr}$

Graphical Analysis — Non-Uniform Motion:

For curved displacement-time graphs:

• The slope at any point gives instantaneous velocity
• To find acceleration, plot the velocity-time graph from the slope of the s-t graph

For curved velocity-time graphs:

• Plot acceleration against time by finding the slope of the v-t graph
• Area under a-t graph gives change in velocity (not displacement)

Average Velocity vs Average Speed:

• Average velocity = total displacement ÷ total time (vector)
• Average speed = total distance ÷ total time (scalar)

For a journey with multiple segments: $$\bar{v} = \frac{s_1 + s_2 + s_3}{t_1 + t_2 + t_3}$$

⚡ NABTEB Quick Reference:

• $v = u + at$
• $s = ut + \frac{1}{2}at^2$
• $v^2 = u^2 + 2as$
• $s = \frac{(u+v)}{2}t$
• Free fall: $u = 0$, $a = g = 9.8 , \text{m/s}^2$
• Projectile (horizontal): $t = \sqrt{\frac{2h}{g}}$, $R = u\sqrt{\frac{2h}{g}}$
• Projectile (angle $\theta$): $T = \frac{2u\sin\theta}{g}$, $R = \frac{u^2\sin 2\theta}{g}$
• Circular: $a_c = \frac{v^2}{r} = \omega^2 r$