Kinematics

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your MDCAT exam.

Key Definitions:

• Distance: Total path length travelled (scalar, ≥ 0)
• Displacement: Change in position (vector, can be negative/positive)
• Speed: Distance/time (scalar)
• Velocity: Displacement/time (vector)
• Acceleration: Rate of change of velocity

Equations of Motion (Constant Acceleration):

$$v = u + at$$

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

$$v^2 = u^2 + 2as$$

$$s = \frac{u+v}{2}t$$

Where:

• $u$ = initial velocity (m/s)
• $v$ = final velocity (m/s)
• $s$ = displacement (m)
• $a$ = acceleration (m/s²)
• $t$ = time (s)

⚡ MDCAT Tip: Choose the right equation based on what’s given. If no time is given, use $v^2 = u^2 + 2as$. If acceleration is zero, $v = u$ and $s = ut$.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

Motion Under Gravity:

For objects in free fall (ignoring air resistance):

• $g = 9.8,\text{m/s}^2$ downward (≈ $10,\text{m/s}^2$ for approximations)
• Upward motion: $a = -g$
• Downward motion: $a = +g$

For upward projection:

• At maximum height: $v = 0$
• $v = u - gt = 0$ at max height
• Time to maximum height $t = u/g$

For freely falling body (dropped from rest, $u = 0$):

• $v = gt$
• $s = \frac{1}{2}gt^2$
• $t = \sqrt{\frac{2s}{g}}$

Worked Example: A ball is thrown vertically upward with $30,\text{m/s}$. Find:

1. Maximum height
2. Time to return to starting point

Solution:

1. Using $v^2 = u^2 - 2gh$ (upward, so $a = -g$): $0 = 900 - 2(10)h$ $h = 45,\text{m}$

2. Time up = $u/g = 30/10 = 3,\text{s}$ Time down = same as time up = $3,\text{s}$ (in vacuum) Total = $6,\text{s}$

Graphical Analysis:

Displacement-Time Graph:

• Gradient = velocity
• Curved line → changing velocity
• Steeper slope → higher velocity

Velocity-Time Graph:

• Gradient = acceleration
• Area under graph = displacement
• Negative area (below time axis) = negative displacement

Acceleration-Time Graph:

• Area = change in velocity

⚡ Common Mistake: Confusing distance with displacement. Distance is the total path length and always positive. Displacement is the shortest path between start and end points and can be negative.

Relative Velocity:

If two objects move in the same direction with velocities $v_A$ and $v_B$: $$v_{A/B} = v_A - v_B$$ (velocity of A relative to B)

If in opposite directions: $$v_{A/B} = v_A + v_B$$

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🔴 Extended — Deep Study (3mo+)

Comprehensive theory for serious exam preparation.

Projectile Motion:

For motion in a plane with constant $g$:

Horizontal component: $a_x = 0$, so $v_x = u\cos\theta$ (constant) $$x = (u\cos\theta)t$$

Vertical component: $a_y = -g$, so $v_y = u\sin\theta - gt$ $$y = (u\sin\theta)t - \frac{1}{2}gt^2$$

At maximum height: $v_y = 0$

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

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

Range: $R = \frac{u^2\sin 2\theta}{g}$

Key Result: For fixed launch speed $u$, maximum range occurs at $\theta = 45°$: $$R_{\max} = \frac{u^2}{g}$$

Worked Example: A projectile is fired at $100,\text{m/s}$ at $30°$ to the horizontal. Find:

1. Time of flight
2. Maximum height
3. Range

Solution:

1. $T = \frac{2(100)\sin 30°}{10} = \frac{200(0.5)}{10} = 10,\text{s}$
2. $H = \frac{100^2 \times (0.5)^2}{20} = \frac{10000 \times 0.25}{20} = 125,\text{m}$
3. $R = \frac{100^2 \times \sin 60°}{10} = \frac{10000 \times 0.866}{10} = 866,\text{m}$

Uniform Circular Motion:

For an object moving in a circle of radius $r$ with constant speed $v$:

• Angular velocity: $\omega = \frac{v}{r} = \frac{2\pi}{T}$
• Period: $T = \frac{2\pi r}{v}$
• Frequency: $f = \frac{1}{T}$

Centripetal Acceleration: $a_c = \frac{v^2}{r} = \omega^2 r$

Centripetal Force: $F_c = \frac{mv^2}{r} = m\omega^2 r$

⚡ MDCAT Pattern:

Year	Question	Concept
2023	Projectile motion	Range/height calculation
2022	Velocity-time graph	Area = displacement
2021	Relative velocity	Moving observer

Non-Uniform Circular Motion:

When speed is changing:

• Tangential acceleration: $a_t = \frac{dv}{dt}$
• Total acceleration: $a = \sqrt{a_c^2 + a_t^2}$
• Direction of $a$ is NOT toward centre

Simple Harmonic Motion (SHM):

For motion where $a = -\omega^2 x$:

• $x = A\cos(\omega t + \phi)$
• $v = -A\omega\sin(\omega t + \phi)$
• $a = -A\omega^2\cos(\omega t + \phi)$
• Period: $T = \frac{2\pi}{\omega}$ (independent of amplitude!)

SHM and Circular Motion: SHM is the projection of uniform circular motion onto one axis.

Velocity in SHM: $$v = \omega\sqrt{A^2 - x^2}$$

Maximum velocity at $x = 0$: $v_{\max} = \omega A$

Acceleration in SHM: $$a = -\omega^2 x$$

Maximum acceleration at $x = ±A$: $a_{\max} = \omega^2 A$

⚡ Exam Strategy: Always start by defining your positive direction. Draw diagrams for projectile problems, showing initial velocity components. For circular motion, remember centripetal force is directed toward the centre — there’s no outward force (centrifugal is a “pseudo-force” in rotating frames).

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