Curriculum

Mathematics

Unit 2

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Unit 5

0/12

Text lesson

Motion with Variable Acceleration and Force

Mathematics

Motion with Variable Acceleration and Force

2-1 (a) Newton’s second law as a differential equation

The law

The rate of change of momentum is proportional to the force and in its direction: $\vec{F} = \frac{d}{d t} (m \vec{v})$ .
When the mass is constant this gives $\vec{F} = m \vec{a}$ with the units chosen so that the constant of proportionality is 1 (newton for $m$ in kg and $a$ in m/s $^{2}$ ; dyne for g and cm/s $^{2}$ ).
1\ kg·wt = 9.8 N and $1 gf = 980$ dyne.

Acceleration is $a = \frac{d v}{d t}$ , and by the chain rule with $v = \frac{d s}{d t}$ we also have
$a = \frac{d v}{d t} = \frac{d v}{d s} \cdot \frac{d s}{d t} = v \frac{d v}{d s} = \frac{d}{d s} (\frac{1}{2} v^{2})$
Which form to use depends on what the force depends on.

Point of comparison	Force depends on time $t$	Force depends on position $s$	Force depends on velocity $v$
Equation used	$m \frac{d v}{d t} = F (t)$	$m v \frac{d v}{d s} = F (s)$	$m \frac{d v}{d t} = F (v)$ or $m v \frac{d v}{d s} = F (v)$
Variables separated	$d v = \frac{F (t)}{m} d t$	$v d v = \frac{F (s)}{m} d s$	$\frac{d v}{F (v)} = \frac{d t}{m}$ or $\frac{v d v}{F (v)} = \frac{d s}{m}$
First answer	$v (t)$	$v (s)$ (speed at a place)	$v (t)$ or $v (s)$
Then	integrate again for $s (t)$	$\frac{d s}{d t} = v (s)$ for $t$	integrate for the other variable

Impulse–momentum∫t1t2F dt=m(v2−v1).

Work–energy∫s1s2F ds=12m(v22−v12).

Constant acceleration is a special casea=c gives v=v0+ct, s=v0t+12ct2 and v2=v02+2cs.

Draw the body and the forces; take the direction of motion as positive.
Write the equation of motion $F_{resultant} = m a$ and choose the form of $a$ .
Separate the variables and integrate between the initial and final values (limits), or integrate indefinitely and find the constant from the initial condition.
Keep units consistent.

Motion with Variable Acceleration and Force · 2 / 12

2-1 (b) Acceleration as a function of time

Method

If $a = f (t)$ , then $v = v_{0} + \int_{0}^{t} f (u) d u$ and $s = s_{0} + \int_{0}^{t} v d t$ .

Motion with Variable Acceleration: a cart on a rail with three light gates shows motion with variable acceleration: position s = t³, velocity 3t², acceleration 6t.

✍️ Worked examples

Example 1:
A particle has acceleration $a = (6 t - 4)$ m/s $^{2}$ and starts from the origin with velocity 2 m/s.
Find $v$ and $s$ at time $t$ and their values at $t = 3$ .
Does the particle ever stop?
Solution:
$v = 2 + \int_{0}^{t} (6 u - 4) d u = 3 t^{2} - 4 t + 2$ .
$s = \int_{0}^{t} (3 u^{2} - 4 u + 2) d u = t^{3} - 2 t^{2} + 2 t$ .
At t=3 v=27−12+2=17 m/s and s=27−18+6=15 m.
v=0 needs 3t2−4t+2=0, with discriminant 16−24<0 there is no solution, so the particle never stops.
Example 2:
A force $F = 12 t - 6$ N acts on a body of mass 3 kg at rest.
Find its velocity after 4 s and the distance travelled in that time (the body moves in a straight line).
Solution:
$a = \frac{F}{3} = 4 t - 2$ .
$v = \int_{0}^{t} (4 u - 2) d u = 2 t^{2} - 2 t$ ; at $t = 4$ , $v = 32 - 8 = 24$ m/s.
$s = \int_{0}^{t} (2 u^{2} - 2 u) d u = \frac{2}{3} t^{3} - t^{2}$ ; at $t = 4$ , $s = \frac{128}{3} - 16 = \frac{80}{3} \approx 26.7$ m.
Note v<0 for 0<t<1, so the body first moves backwards; the displacement is not the distance if the direction reverses.
Example 3:
A force $F = 6 + 12 t$ N acts on a particle of mass 2 kg which has velocity 3 m/s at $t = 0$ .
Find the impulse during the interval $[1, 3]$ and the velocity at $t = 3$ .
Solution:
Impulse $= \int_{1}^{3} (6 + 12 t) d t = {[6 t + 6 t^{2}]}_{1}^{3} = (18 + 54) - (6 + 6) = 60$ N·s.
Velocity at t=1 v1=3+12∫01(6+12t) dt=3+12(6+6)=9 m/s.
Velocity at t=3 2v3=2v1+60, so v3=9+30=39 m/s.

Change of momentum

The change of momentum in $[t_{1}, t_{2}]$ is $Δ H = m \int_{t_{1}}^{t_{2}} a d t$ .

✍️ Worked examples

Example 4 A car of mass 1.5 tonnes has acceleration $a = 12 t - t^{2}$ m/s $^{2}$ . Find the change of momentum in the first 6 s and in the interval $[2, 14]$ .
Solution:
$m = 1500$ kg. An antiderivative of $a$ is $6 t^{2} - \frac{t^{3}}{3}$ .
First 6 s 6(36)−2163=216−72=144, so ΔH=1500×144=216 000 kg·m/s.
Interval [2,14] at 14: 6(196)−27443=1176−914.67=261.33; at 2: 24−83=21.33.
Difference $= 240$ , so $Δ H = 1500 \times 240 = 360 000$ kg·m/s.
Example 5 A body of mass 8 kg has $a = 2 t - 6$ . Find the change of momentum between $t = 3$ and $t = 5$ .
Solution:
$Δ H = 8 \int_{3}^{5} (2 t - 6) d t = 8 {[t^{2} - 6 t]}_{3}^{5} = 8 [(25 - 30) - (9 - 18)] = 8 (4) = 32$ kg·m/s.

Motion with Variable Acceleration and Force · 3 / 12

2-1 (c) Acceleration as a function of position

Method

If the force depends on the distance $r$ from a fixed point, use $m v \frac{d v}{d r} = F (r)$ .
Separate ∫v0vv dv=1m∫r0rF dr, i.e. 12(v2−v02)=1m∫F dr.
To find the time afterwards, use $v = \frac{d r}{d t}$ and integrate $\frac{d r}{v}$ .

✍️ Worked examples

Example 6:
A body of mass 2 kg starts from rest at the origin under the force $F = (8 - 2 x)$ N along the $x$ -axis ( $x$ in metres).
Find $v$ in terms of $x$ , the greatest speed, and where the body first stops.
Solution:
$a = \frac{8 - 2 x}{2} = 4 - x$ , so $v \frac{d v}{d x} = 4 - x$ .
$\int_{0}^{v} v d v = \int_{0}^{x} (4 - x) d x$ gives $\frac{v^{2}}{2} = 4 x - \frac{x^{2}}{2}$ , so $v^{2} = 8 x - x^{2}$ .
The speed is greatest where $a = 0$ , i.e. $x = 4$ m: $v^{2} = 32 - 16 = 16$ , so $v_{m a x} = 4$ m/s.
The body stops when v=0 x(8−x)=0, so x=8 m. There the force is −8 N, so it turns back.
Example 7:
A body of mass 1 kg is released from rest at the origin and acted on by $F = 5 r + 6$ N, with $r$ in metres.
Find $v$ when $r = 4$ m, and $r$ when $v = 9$ m/s.
Solution:
$v \frac{d v}{d r} = 5 r + 6$ , so $\frac{v^{2}}{2} = \frac{5}{2} r^{2} + 6 r$ (the constant is 0 since $v = 0$ at $r = 0$ ).
At r=4 v22=40+24=64, so v2=128 and v=82≈11.3 m/s.
At v=9 812=52r2+6r, so 5r2+12r−81=0 and (5r+27)(r−3)=0.
$r = 3$ m (the root $- \frac{27}{5}$ is rejected).
Example 8:
A particle of mass 1 kg moves along a line under an attractive force towards the origin of magnitude $4 x$ N at distance $x$ .
At $x = 0$ its speed is 6 m/s.
Show that it stops at $x = \pm 3$ m.
Solution:
The force is towards the origin, so a=−4x vdvdx=−4x.
$\frac{v^{2}}{2} = \frac{36}{2} - 2 x^{2}$ , so $v^{2} = 36 - 4 x^{2}$ .
$v = 0$ when $x^{2} = 9$ , i.e. $x = \pm 3$ m.
The motion is an oscillation between $- 3$ and $3$ (simple harmonic with period $\frac{2 π}{2} = π$ s).
Example 9 A body falls from rest with acceleration $a = g (1 - \frac{x}{L})$ where $x$ is the distance fallen. (Take $g = 9.8$ and $L = 49$ m.) Find its speed when $x = 24.5$ m.
Solution:
$v \frac{d v}{d x} = g (1 - \frac{x}{L})$ , so $\frac{v^{2}}{2} = g (x - \frac{x^{2}}{2 L})$ .
At $x = 24.5$ : $\frac{x^{2}}{2 L} = \frac{600.25}{98} = 6.125$ , so $\frac{v^{2}}{2} = 9.8 (24.5 - 6.125) = 9.8 \times 18.375 = 180.075$ .
$v^{2} = 360.15$ , so $v \approx 18.98$ m/s.

Motion with Variable Acceleration and Force · 4 / 12

2-1 (d) Acceleration as a function of velocity

Method

If a=f(v), separate in time dvf(v)=dt, or in distance: v dvf(v)=ds.
A resistance proportional to the velocity gives $a = - k v$ and an exponential decay.

✍️ Worked examples

Example 10:
A boat of mass $m$ moving at $v_{0}$ experiences a resistance $k v$ (where $k$ is constant) and no other horizontal force.
Show that $v = v_{0} e^{- k t / m}$ and that the distance travelled before it is brought to rest is $\frac{m v_{0}}{k}$ .
Solution:
$m \frac{d v}{d t} = - k v$ , so $\int_{v_{0}}^{v} \frac{d v}{v} = - \frac{k}{m} \int_{0}^{t} d t$ .
$\ln \frac{v}{v_{0}} = - \frac{k t}{m}$ , so $v = v_{0} e^{- k t / m}$ .
Using $v \frac{d v}{d s} = - \frac{k v}{m}$ we get $d v = - \frac{k}{m} d s$ , so $v = v_{0} - \frac{k}{m} s$ .
$v = 0$ when $s = \frac{m v_{0}}{k}$ . The velocity tends to zero only as $t \to \infty$ , but the distance is finite.
Example 11:
A body of mass 3 kg moves in a line with initial velocity 2 m/s under the force $F = \frac{18}{2 v + 1}$ N, $v$ being its velocity.
Find the time to reach 6 m/s and the distance travelled meanwhile.
Solution:
$a = \frac{F}{3} = \frac{6}{2 v + 1}$ , so $(2 v + 1) d v = 6 d t$ .
Integrate v2+v=6t+C. At t=0, v=2: C=6.
At v=6 36+6=6t+6, so t=6 s.
For the distance use $v \frac{d v}{d s} = \frac{6}{2 v + 1}$ , so $v (2 v + 1) d v = 6 d s$ .
$6 s = {[\frac{2}{3} v^{3} + \frac{v^{2}}{2}]}_{2}^{6} = (144 + 18) - (\frac{16}{3} + 2) = 162 - \frac{22}{3} = \frac{464}{3}$ .
$s = \frac{464}{18} = \frac{232}{9} \approx 25.8$ m.
Example 12:
A car of mass 1000 kg has an engine force $F = 3000$ N and resistance $R = 50 v$ N.
Starting from rest, find $v$ in terms of $t$ and its limiting speed.
Solution:
$1000 \frac{d v}{d t} = 3000 - 50 v$ , so $\frac{d v}{60 - v} = \frac{d t}{20}$ .
Integrate −ln⁡(60−v)=t20+C. At t=0, v=0 gives C=−ln⁡60.
$\ln \frac{60}{60 - v} = \frac{t}{20}$ , so $v = 60 (1 - e^{- t / 20})$ .
As $t \to \infty$ the speed approaches the limiting speed $60$ m/s (where driving force equals resistance).

Motion with Variable Acceleration and Force · 5 / 12

2-1 (e) Vector form of the equation of motion

Method

If $\vec{F}$ and $\vec{S}$ are given as vectors in $\hat{i}, \hat{j}, \hat{k}$ , then $\vec{v} = \frac{d \vec{S}}{d t}$ , $\vec{a} = \frac{d \vec{v}}{d t}$ and $\vec{F} = m \vec{a}$ component by component.
To find $\vec{v}$ and $\vec{S}$ from $\vec{F} (t)$ , integrate each component separately, using the initial conditions.
If the resultant force is zero, the acceleration is zero and the velocity is constant (Newton’s first law).

✍️ Worked examples

Example 13 A body of mass 2 kg moves under three forces ${\vec{F}}_{1} = a \hat{i} + \hat{j}$ , ${\vec{F}}_{2} = 2 \hat{i} - 3 \hat{j}$ , ${\vec{F}}_{3} = 3 \hat{i} + b \hat{j}$ N, with displacement $\vec{S} = (t^{2} + 1) \hat{i} + 3 t^{2} \hat{j}$ m. Find $a$ and $b$ .
Solution:
$\vec{v} = 2 t \hat{i} + 6 t \hat{j}$ and $\vec{a} = 2 \hat{i} + 6 \hat{j}$ .
$\vec{F} = m \vec{a} = 4 \hat{i} + 12 \hat{j}$ .
Sum of the forces (a+5)i^+(b−2)j^.
$a + 5 = 4$ gives $a = - 1$ ; $b - 2 = 12$ gives $b = 14$ .
Example 14 A body moving with $\vec{S} = 2 t \hat{i} - t \hat{j} + \hat{k}$ is acted on by ${\vec{F}}_{1} = 4 \hat{i} + 3 \hat{k}$ , ${\vec{F}}_{2} = - \hat{i} + 4 \hat{j} - 15 \hat{k}$ and ${\vec{F}}_{3}$ . Find $∥ {\vec{F}}_{3} ∥$ .
Solution:
$\vec{v} = 2 \hat{i} - \hat{j}$ is constant, so $\vec{a} = \vec{0}$ and the resultant force is zero.
${\vec{F}}_{3} = - {\vec{F}}_{1} - {\vec{F}}_{2} = (- 4 + 1) \hat{i} + (0 - 4) \hat{j} + (- 3 + 15) \hat{k} = - 3 \hat{i} - 4 \hat{j} + 12 \hat{k}$ .
$∥ {\vec{F}}_{3} ∥ = \sqrt{9 + 16 + 144} = 13$ force units.
Example 15 A force $\vec{F} = (6 t + 4) \hat{i} + 12 t^{2} \hat{j}$ N acts on a body of mass 2 kg at rest at the origin. Find $\vec{v}$ , $\vec{S}$ and the speed at $t = 2$ .
Solution:
$\vec{a} = (3 t + 2) \hat{i} + 6 t^{2} \hat{j}$ .
$\vec{v} = (\frac{3}{2} t^{2} + 2 t) \hat{i} + 2 t^{3} \hat{j}$ .
$\vec{S} = (\frac{1}{2} t^{3} + t^{2}) \hat{i} + \frac{1}{2} t^{4} \hat{j}$ .
At t=2 v⃗=(6+4)i⁺¹⁶j^=10i⁺¹⁶j^ and the speed is 100+256=356≈18.9 m/s.

Motion with Variable Acceleration and Force · 6 / 12

2-1 (f) Momentum and a variable mass

Rules

The momentum is $\vec{H} = m \vec{v}$ and the force is $\vec{F} = \frac{d \vec{H}}{d t} = m \frac{d \vec{v}}{d t} + \vec{v} \frac{d m}{d t}$ .
If the velocity is constant and the mass changes, $F = v \frac{d m}{d t}$ .
If the mass is constant, $F = m a$ .
A constant rate of gain or loss gives $m = m_{0} + \frac{d m}{d t} t$ .

✍️ Worked examples

Example 16 A body of mass $m = (2 + t)$ kg moves with velocity $v = 3 t$ m/s along a line. Find the momentum and the force at $t = 2$ .
Solution:
$H = m v = (2 + t) (3 t) = 6 t + 3 t^{2}$ .
$F = \frac{d H}{d t} = 6 + 6 t$ ; at $t = 2$ , $F = 18$ N.
Check by the product rule mdvdt+vdmdt=(4)(3)+(6)(1)=18 N.
Example 17:
A metal ball of mass 100 g moves at constant 10 m/s through dusty air; dust sticks to it at 0.6 g/s.
Find its mass and the force on it at time $t$ .
Solution:
Mass m=100+0.6t g.
Force F=vdmdt=(1000 cm/s)(0.6 g/s)=600 dyne, constant.
Example 18 The displacement of a body of variable mass $m = (4 t + 1)$ g is $S = (t^{2} + 2 t)$ cm. Find the momentum and the force at $t = 4$ .
Solution:
$v = 2 t + 2$ ; $H = (4 t + 1) (2 t + 2) = 8 t^{2} + 10 t + 2$ g·cm/s.
$F = \frac{d H}{d t} = 16 t + 10$ ; at $t = 4$ , $F = 74$ dyne.

Motion with Variable Acceleration and Force · 7 / 12

2-1 (g) Apparent weight with changing acceleration

Idea

A person in a lift presses on the floor with force $N = m (g + a)$ when the lift accelerates upward with acceleration $a$ .
If $a$ varies with time, so does $N$ ; just replace $a$ by the function.

✍️ Worked examples

Example 19:
A lift accelerates upwards with $a = 0.2 t$ m/s $^{2}$ .
A man of mass 60 kg stands in it.
Find the pressure on the floor at $t = 5$ s in kg·wt, and the speed of the lift at that time (starting from rest).
Solution:
At t=5 a=1 m/s2.
$N = 60 (9.8 + 1) = 648$ N $= \frac{648}{9.8} \approx 66.1$ kg·wt.
$v = \int_{0}^{5} 0.2 t d t = 0.1 (25) = 2.5$ m/s.

Common mistakes

Using $s = u t + \frac{1}{2} a t^{2}$ when the acceleration is not constant.
Forgetting the constant of integration, or using the wrong initial condition.
Mixing the forms $a = \frac{d v}{d t}$ and $a = v \frac{d v}{d s}$ .
Confusing displacement with distance when the velocity changes sign.
Mixing units (grams with newtons, or cm with m).

Motion with Variable Acceleration and Force · 8 / 12

Comparison aspect	Main idea	Keywords
Fundamental Speed Equation	[ الصيغة: v = d / t \| d = v * t \| t = d / v (Units: m/s or km/h conversion factor: 1 km/h = 5/18 m/s). ]	الصيغةUnitsconversionfactor
Relative Speed Rules	[ الصيغة: 1) Stationary observer: Relative Speed = Real Speed	الصيغةStationaryobserverReal
2) Opposite direction	Relative Speed = Real Speed + Observer Speed (appears faster)	RelativeSpeedRealObserver
3) Same direction	Relative Speed = Real Speed – Observer Speed (appears slower)	RelativeSpeedRealObserver
4) Same direction and same speed	Relative Speed = 0 (appears at rest). ]	Relativeappearsrest
Uniform Acceleration Formula	[ الصيغة: a = (v2 – v1) / Δt (where v1 is initial velocity v2 is final velocity and Δt is time elapsed). Positive acceleration: v2 > v1 Negative acceleration (deceleration): v2 < v1 Zero acceleration: v1 = v2 (moving at constant uniform speed). ]	الصيغةinitialvelocityfinal

🔍 In-Depth Explanations, Deductions & Multi-Book Insights:

★ كتاب الشرح والتفصيل العلمي الشامل المعتمد ★

كتاب الشرح المفصل→Motion in One Direction, Speed, Graphical Representation, and Uniform Acceleration

| | 2026/2027 |

🎯 أهداف التعلم والمخرجات التعليمية المقررة بالتفصيل:
• Define motion in one direction along a straight line as the simplest form of motion.
• Derive and calculate regular speed, irregular speed, average speed, and relative speed under different observer conditions.
• Graphically represent uniform speed using distance-time and speed-time Cartesian graphs.
• Formulate uniform acceleration (a = (Δ v) / (Δ t)), distinguishing between accelerating (positive) and decelerating (negative) motion.

◄ 1. Mechanics of Motion in One Direction & Speed Classification
Motion is the most ubiquitous physical phenomenon in our universe.
In classical Newtonian kinematics, motion in one direction along a straight line represents the fundamental baseline for all mechanical systems, exemplified by metro transit systems and railway trains travelling between fixed stations.
The rate of motion depends strictly on two independent physical variables: the spatial trajectory distance (d) and the elapsed temporal duration (t).
In real-world traffic dynamics, vehicles almost never maintain a perfectly uniform speed due to road conditions, topography, traffic signals, and variable congestion.
Therefore, physics introduces the concept of average speed (v̄), which represents the hypothetical constant velocity that would cover the identical gross distance over the identical total timeframe.

شكل توضيحي (figure_diagram_1.png): شكل ومخطط تحليلي معتمد لدرس: Motion in One Direction
Speed
Graphical Representation
and Uniform Acceleration

◄ 2. Graphical Modeling & Acceleration Dynamics

Graphical analysis converts kinematic data into visual geometric functions.
On a distance-time (d-t) graph, uniform speed yields a straight linear incline passing through the origin (slope = v).
On a speed-time (v-t) graph, uniform speed is depicted as a horizontal line parallel to the time axis (slope = a = 0).
When an object undergoes acceleration, its velocity vector changes over time.
If a vehicle starts from rest (v1 = 0), its acceleration is strictly positive.
Conversely, when a driver applies the mechanical brakes, the car decelerates until it reaches complete standstill (v2 = 0), resulting in a negative acceleration sign.

🧮 التطبيقات والمسائل النموذجية المحلولة بالخطوات (Step-by-Step Worked Examples):
مثال نموذجي 1:
نص المسألة / السؤال A racing sports car starts from rest and reaches a velocity of 90 km/h in 5 seconds. Later, the driver steps on the brake pedal to avoid an obstacle, bringing the car to a full stop in 2.5 seconds. Calculate: 1) The acceleration during the first stage in m/s2. 2) The deceleration during braking in m/s2.
خطوات الحل والشرح النموذجي Step 1: Convert 90 km/h to m/s: v2 = 90 * (5/18) = 25 m/s. Initial velocity v1 = 0 m/s. Elapsed time Δt1 = 5 s.
Acceleration a1 = (v2 – v1) / Δt1 = (25 – 0) / 5 = +5 m/s² (Uniform positive acceleration).

Braking stage: Initial velocity v1 = 25 m/s, final velocity v2 = 0 m/s, Δt2 = 2.5 s.
Acceleration a2 = (v2 – v1) / Δt2 = (0 – 25) / 2.5 = -10 m/s² (Uniform negative acceleration / deceleration).
💡 أسرار وتكات التفوق وملاحظات خبير المادة (Expert Notes & Exam Secrets):
★ ('Unit Conversion Trap (km/h vs m/s)', 'Never plug km/h directly into an acceleration formula with time in seconds!
You must multiply by 5/18 first to convert km/h to m/s.')
"Calculate the deceleration", and you write "Deceleration = -10 m/s2", it is a double negative trap. You should write: "Acceleration = -10 m/s2" or "Deceleration = 10 m/s2".')

★ ('Zero Acceleration Condition', 'When an object moves at a constant uniform speed, its acceleration is strictly ZERO because Δv = 0.')
🌍 التطبيقات الحياتية والصناعية والتكنولوجية (Real-World Applications):
✔ ('Automotive Safety & Advanced Driver Assistance Systems (ADAS)', 'Modern autonomous vehicles and radar-guided cruise control systems compute relative speed and braking acceleration in real time using high-frequency Doppler radar to maintain safe headway distances.')
✔ ('High-Speed Electric Rail Network (Egypt Monorail & High-Speed Train)', 'The Egyptian national high-speed train project utilizes precise kinematic acceleration and deceleration modeling to optimize power consumption and passenger comfort during transit.')

Motion with Variable Acceleration and Force · 12 / 12

Curriculum

Mathematics

Unit 1

Unit 2

Unit 3

Unit 4

Unit 5

Unit 6

Unit 7

Motion with Variable Acceleration and Force

🌳 Concept tree

Motion with Variable Acceleration and Force

│ ├── The law

│ ├── Three forms of the acceleration

│ ├── Integrals that follow

│ └── General method

├── 2-1 (b) Acceleration as a function of time

│ ├── Method

│ └── Change of momentum

│ └── Method

│ └── Method

│ └── Method

├── 2-1 (f) Momentum and a variable mass

│ └── Rules

├── Idea

└── Common mistakes

The law

Method

✍️ Worked examples

Change of momentum

✍️ Worked examples

Method

✍️ Worked examples

Method

✍️ Worked examples

Method

✍️ Worked examples

Rules

✍️ Worked examples

Idea

✍️ Worked examples

Common mistakes

★ كتاب الشرح والتفصيل العلمي الشامل المعتمد ★

| | 2026/2027 |

◄ 2. Graphical Modeling & Acceleration Dynamics

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