The derivative of $f(x) = 5^{x^2}$ with respect to $x$ is:

B) $2x \cdot 5^{x^2} \cdot \ln 5$

The sum of the first n terms of an arithmetic sequence (S_n) is formulated as:

S_n = n / 2 * [ 2a + (n - 1) * d ] = n / 2 * [ a + l ]

The sum to infinity (S_infinity) of a geometric series exists and converges if and only if the common ratio satisfies:

|r| < 1 (i.e. -1 < r < 1), and S_infinity = a / (1 - r)

ما هو قانون نيوتن الثاني رياضياً؟ What is Newton Second Law formula?

الصيغة الرياضية لقانون نيوتن الثاني هي F = ma (حيث F القوة المحصلة بالنيوتن، m الكتلة بالكيلوجرام، a التسارع بـ m/s²). وعند حساب الوزن: W = mg. [الصيغة الرمزية الموزونة: F = m × a | W = m × g | F₁₂ = -F₂₁]

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Motion with Variable Acceleration and Force

Q: ما هو قانون نيوتن الثاني رياضياً؟ What is Newton Second Law formula?

الصيغة الرياضية لقانون نيوتن الثاني هي F = ma (حيث F القوة المحصلة بالنيوتن، m الكتلة بالكيلوجرام، a التسارع بـ m/s²). وعند حساب الوزن: W = mg. [الصيغة الرمزية الموزونة: F = m × a | W = m × g | F₁₂ = -F₂₁]

Course: Mathematics Unit: Secondary 3 Mathematics 2026/2027 — further lessons

🧭 Lesson points — pick a point and study it with its question bank

1. 2-1 (a) Newton's second law as a differential equationExplanation

2. 2-1 (b) Acceleration as a function of timeExplanation

3. 2-1 (c) Acceleration as a function of positionExplanation

4. 2-1 (d) Acceleration as a function of velocityExplanation

5. 2-1 (e) Vector form of the equation of motionExplanation

6. 2-1 (f) Momentum and a variable massExplanation

7. 2-1 (g) Apparent weight with changing accelerationExplanation

This lesson develops lesson 2-1 of the dynamics unit "Applications on Newton's Laws of Motion" using calculus.
After it you should be able to:

write Newton's second law in three differential forms and choose the right one;
solve motion in a straight line when the force or acceleration depends on time, position or velocity;
work with vector forces, momentum change and a mass that varies with time.

🌳 Concept tree

Motion with Variable Acceleration and Force
├── 2-1 (a) Newton&#x27;s second law as a differential equation
│   ├── 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
├── 2-1 (c) Acceleration as a function of position
│   └── Method
├── 2-1 (d) Acceleration as a function of velocity
│   └── Method
├── 2-1 (e) Vector form of the equation of motion
│   └── Method
├── 2-1 (f) Momentum and a variable mass
│   └── Rules
└── 2-1 (g) Apparent weight with changing acceleration
    ├── Idea
    └── Common mistakes

01 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\ \text{kg·wt} = 9.8 N and $1 gf = 980$ dyne.

Three forms of the acceleration

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

Integrals that follow

Impulse–momentum: $\int_{t_{1}}^{t_{2}} F d t = m (v_{2} - v_{1})$ .
Work–energy: $\int_{s_{1}}^{s_{2}} F d s = \frac{1}{2} m (v_{2}^{2} - v_{1}^{2})$ .
Constant acceleration is a special case: $a = c$ gives $v = v_{0} + c t$ , $s = v_{0} t + \frac{1}{2} c t^{2}$ and $v^{2} = v_{0}^{2} + 2 c s$ .

General method

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.

02 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 $s = s_{0} + \int_{0}^{t} v d t$ .

✍️ 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 $3 t^{2} - 4 t + 2 = 0$ , with discriminant $16 - 24 < 0$ : there is no solution, so the particle never stops.

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: $6 (36) - \frac{216}{3} = 216 - 72 = 144$ , so $Δ H = 1500 \times 144 = 216 000$ kg·m/s.
Interval $[2, 14]$ : at $14$ : $6 (196) - \frac{2744}{3} = 1176 - 914.67 = 261.33$ ; at $2$ : $24 - \frac{8}{3} = 21.33$ .
Difference $= 240$ , so $Δ H = 1500 \times 240 = 360 000$ kg·m/s.

03 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: $\int_{v_{0}}^{v} v d v = \frac{1}{m} \int_{r_{0}}^{r} F d r$ , i.e. $\frac{1}{2} (v^{2} - v_{0}^{2}) = \frac{1}{m} \int F d r$ .
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.

04 2-1 (d) Acceleration as a function of velocity

Method

If $a = f (v)$ , separate in time: $\frac{d v}{f (v)} = d t$ , or in distance: $\frac{v d v}{f (v)} = d s$ .
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.

05 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: $(a + 5) \hat{i} + (b - 2) \hat{j}$ .
$a + 5 = 4$ gives $a = - 1$ ; $b - 2 = 12$ gives $b = 14$ .

06 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: $m \frac{d v}{d t} + v \frac{d m}{d t} = (4) (3) + (6) (1) = 18$ N.

07 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/s $^{2}$ .
$N = 60 (9.8 + 1) = 648$ 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).

📖

📚 Comprehensive Multi-Textbook Encyclopedia & Key Terminology

📌 Comprehensive Scientific Terminology Dictionary:

Motion (الحركة): The continuous change in an object's position over time relative to a fixed reference point. Motion in a straight line in one direction is the simplest type of motion.

Uniform Speed (السرعة المنتظمة): The speed at which an object covers equal distances in equal intervals of time, regardless of how small these time intervals are.

Average Speed (السرعة المتوسطة: v̄): The total distance traveled by a moving object divided by the total time taken to cover that distance: $v_avg = \frac{d_total}{t_total}$.

Relative Speed (السرعة النسبية): The speed of a moving object relative to a specific observer, which changes depending on whether the observer is stationary or moving in the same or opposite direction.

Uniform Acceleration (العجلة المنتظمة: a): The acceleration in which the speed of an object changes (increases or decreases) by equal amounts in equal intervals of time: $a = \frac{v₂ - v₁}{\Delta t}$, measured in $m/s²$.

⚖️ Governing Physical Laws & Mathematical Relations:

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). ]

Relative Speed Rules: [ الصيغة: 1) Stationary observer: Relative Speed = Real Speed. 2) Opposite direction: Relative Speed = Real Speed + Observer Speed (appears faster). 3) Same direction: Relative Speed = Real Speed - Observer Speed (appears slower). 4) Same direction and same speed: Relative Speed = 0 (appears at rest). ]

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). ]

🔍 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 ($a = \frac{\Delta v}{\Delta t}$), 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_diagram₁.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). Step 2: 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.')

★ ('Sign of Deceleration', 'When a question asks: "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.')