Dynamics | NEB Class 11 Mathematics Notes

Chapter Overview

This chapter is the analytical heart of mechanics — examining how force vectors create physical acceleration and govern the trajectories of moving bodies. Dynamics extends kinematics by directly applying Newton's laws of motion to predict exactly how physical bodies behave over space and time under external influences. In NEB Grade 11 Physics exams, Dynamics is a foundational and high-scoring unit, presenting challenges that range from 1-mark calculus-based MCQs to complex 5-mark long questions involving vertical projections, multi-stage vehicle decelerations, and smooth inclined planes. Master this chapter, and you build the core mechanical intuition required for all of engineering physics.

What this chapter covers

Kinematics vs. Kinetics PrinciplesDisplacement, Velocity, and Acceleration VectorsCalculus-Based Definitions of Motion DerivativesAnalytical Derivations of Equations of MotionDistance Covered in the Specific t-th SecondMotion Under Gravity (Free Fall and Projection)Time of Ascent, Descent, and Flight EquationsMotion on Smooth Inclined PlanesTwo-Stage Acceleration and Braking SystemsRamps and Gravity Component Resolutions

Learning Objectives

1

Differentiate between scalar distance and vector displacement in diverse coordinate configurations

2

Derive the equations of motion ($v=u+at$, $s=ut+1/2at^2$, $v^2=u^2+2as$) using graphical, algebraic, and calculus methods

3

Solve complex kinematics problems involving the specific distance covered in the $t$-th second

4

Analyze vertical motion under gravity, applying correct sign conventions for ascending and descending bodies

5

Formulate and evaluate equations of motion for bodies sliding down or projected up smooth inclined planes

6

Deconstruct multi-stage motion problems into separate intervals of uniform acceleration and deceleration

Calculus Connects Position, Velocity, and Acceleration

Key Insight

In advanced physics, motion is not merely a set of static formulas but a continuous relationship of time derivatives. Position ($s$), velocity ($v$), and acceleration ($a$) are fundamentally linked:

• Velocity is the first derivative of position with respect to time: $v = \frac{ds}{dt}$
• Acceleration is the first derivative of velocity, or the second derivative of position: $a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$

By reversing these derivatives through definite integration, we can derive every kinematic formula for constant acceleration directly, bypassing geometric approximations.

In NEB exams: When a problem states an object moves along an 'inclination of 1 in n', it is a mathematical code indicating that $\sin \theta = \frac{1}{n}$. Use this relationship directly to resolve the gravitational acceleration component.

Key Definitions

Dynamics

The branch of physical mechanics that investigates the laws governing the motion of material bodies under the direct action of forces.

Kinematics

A sub-branch of dynamics that describes the motion of points, bodies, and systems of bodies without considering the forces that cause the motion.

Kinetics

A sub-branch of dynamics that relates the action of forces on bodies to the resulting mass accelerations.

Displacement (s)

A vector quantity representing the shortest straight-line distance from the initial position to the final position of a moving particle.

$$\vec{s} = \vec{r}_{final} - \vec{r}_{initial}$$

Acceleration (a)

The time rate of change of velocity. It is a vector quantity and points in the direction of the net change in velocity.

$$a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$$

Retardation

Negative acceleration, occurring when the magnitude of a body's velocity decreases over time (deceleration).

Free Fall

The state of motion of a body falling under the sole influence of Earth's gravitational pull, with all resistance forces like air drag neglected.

Inclined Plane

A flat surface tilted at an angle $\theta$ to the horizontal, used to reduce the effective gravitational acceleration acting along the path of motion.

$$a = g \sin \theta$$

Distance vs Displacement

Differentiating scalar path properties from vector position changes is essential for correct kinematics setups.

Distance

$$d = \int |v(t)| dt$$

A scalar quantity representing the absolute length of the actual path traversed

Always positive or zero, regardless of the direction of motion

Depends entirely on the specific path taken between two coordinates

Cannot decrease over time for a moving particle

Keywords: 'total path length', 'odometer reading', 'traveled distance'

VS

Displacement

$$\vec{s} = \int \vec{v}(t) dt$$

A vector quantity representing the shortest straight-line path from start to end

Can be positive, negative, or zero, depending on the coordinate direction

Independent of the actual path taken; depends only on the endpoints

Can decrease as a particle returns toward its starting coordinate

Keywords: 'shortest distance', 'position vector change', 'net shift'

Quick check: If an athlete runs one full lap around a 400m circular track, the distance covered is 400m, but their net displacement is exactly 0m.

Complete Formula Sheet with Derivations

Velocity-Time Relation

$$v = u + at$$

u is initial velocity, v is final velocity, a is constant acceleration, t is time.

1We define instantaneous acceleration as the derivative of velocity: $a = \frac{dv}{dt}$.

2Rearrange the differential equation: $dv = a \, dt$.

3Integrate both sides of the equation. Since acceleration $a$ is constant, it can be factored out of the integral: $$\int_{u}^{v} dv = a \int_{0}^{t} dt$$

4Evaluate the definite integrals: $[v]_u^v = a[t]_0^t \Rightarrow v - u = at$.

5Isolate final velocity: $$v = u + at$$

Displacement-Time Relation

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

s is the net displacement accumulated over time interval t.

1We define instantaneous velocity as the derivative of position: $v = \frac{ds}{dt} \Rightarrow ds = v \, dt$.

2Substitute the expression for $v$ derived from the first equation of motion ($v = u + at$): $ds = (u + at) \, dt$.

3Integrate both sides from the initial boundary conditions ($s=0$ at $t=0$) to the final conditions ($s$ at $t$): $$\int_{0}^{s} ds = \int_{0}^{t} (u + at) \, dt$$

4Split the integral on the right side and integrate term-by-term: $s = \int_{0}^{t} u \, dt + \int_{0}^{t} at \, dt$.

5Evaluate the integrals: $$s = ut + \frac{1}{2}at^2$$

Velocity-Displacement Relation

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

Useful when time t is not given or required by the problem.

1We start with the acceleration definition $a = \frac{dv}{dt}$ and apply the chain rule of calculus: $$a = \frac{dv}{ds} \cdot \frac{ds}{dt}$$

2Since $\frac{ds}{dt} = v$, substitute this into the expression: $a = v \frac{dv}{ds}$.

3Rearrange the differential terms to separate variables: $a \, ds = v \, dv$.

4Integrate both sides, taking $a$ outside since it is constant: $$\int_{0}^{s} a \, ds = \int_{u}^{v} v \, dv \Rightarrow a \int_{0}^{s} ds = \int_{u}^{v} v \, dv$$

5Evaluate the integrals: $a[s]_0^s = \left[ \frac{v^2}{2} \right]_u^v \Rightarrow as = \frac{v^2 - u^2}{2}$.

6Isolate final velocity squared: $$v^2 = u^2 + 2as$$

Distance Covered in the Specific t-th Second

s_t = u + \frac{1}{2}a(2t - 1)

This formula gives the displacement during a single specific second t, not the total displacement.

1The displacement in the $t$-th second is defined as the difference between the total displacement in $t$ seconds and the total displacement in $(t-1)$ seconds: $$s_t = s(t) - s(t-1)$$

2Apply the displacement-time relation: $s(t) = ut + \frac{1}{2}at^2$.

3Write the expression for $s(t-1)$: $s(t-1) = u(t-1) + \frac{1}{2}a(t-1)^2 = ut - u + \frac{1}{2}a(t^2 - 2t + 1)$.

4Expand and simplify: $s(t-1) = ut - u + \frac{1}{2}at^2 - at + \frac{1}{2}a$.

5Subtract the two expressions: $s_t = \left( ut + \frac{1}{2}at^2 \right) - \left( ut - u + \frac{1}{2}at^2 - at + \frac{1}{2}a \right)$.

6Cancel common terms: $s_t = u + at - \frac{1}{2}a$.

7Factor out $\frac{1}{2}a$: $$s_t = u + \frac{1}{2}a(2t - 1)$$

Motion on a Smooth Inclined Plane

a = g \sin \theta

θ is the angle of inclination of the plane with the horizontal floor.

1Consider a block of mass $m$ sliding down a smooth plane tilted at angle $\theta$.

2The gravitational force acting vertically downwards is $mg$.

3Resolve the gravitational force vector $mg$ into two perpendicular components:
• Normal to the incline: $mg \cos \theta$
• Parallel to the incline (downward direction): $mg \sin \theta$

4The normal force $mg \cos \theta$ is balanced by the plane's perpendicular normal reaction $R$.

5The net unbalanced force causing motion down the incline is: $F_{\text{net}} = mg \sin \theta$.

6By Newton's second law, $F_{\text{net}} = ma \Rightarrow ma = mg \sin \theta$.

7Divide by $m$ to find the acceleration: $$a = g \sin \theta$$

Kinematic Equation Selection Flowchart

Use this decision tree to identify the correct kinematic equation based on given variables.

Does the problem ask for or provide the displacement in a single SPECIFIC second (e.g., 5th second)?

Yes

Use the t-th second formula: s_t = u + 1/2 a(2t - 1)

No

Continue ↓

Is the TIME variable (t) neither given nor required by the problem?

Yes

Use the velocity-displacement relation: v² = u² + 2as

No

Continue ↓

Is the DISPLACEMENT variable (s or h) neither given nor required?

Yes

Use the velocity-time relation: v = u + at

No

Use the displacement-time relation: s = ut + 1/2 at²

Gravitational Acceleration Sign Convention

A consistent coordinate system prevents algebraic sign errors when solving vertically projected motion problems.

$$\text{Upward Projection: } a = -g \quad \text{and} \quad \text{Downward Falling: } a = +g$$

Why it works (Derivation):

1In vertical motion, the Earth's gravitational acceleration $g$ always pulls vertically downwards toward the center of the Earth.

2To establish a consistent frame of reference, define the vertically upward direction as positive (+y) and the downward direction as negative (-y).

3• Vertically Upward Projection: The initial velocity $u$ is positive (+u). Since gravity pulls downwards, acceleration $a = -g$. The velocity decreases until it reaches zero at maximum height.

4• Vertically Downward Fall (Dropped): The initial velocity $u = 0$, and displacement is directed downwards. Since gravity pulls downwards, we can define the downward direction as positive, setting acceleration $a = +g$.

Example: An object thrown vertically upwards with $u = 30 \text{ ms}^{-1}$ reaches maximum height ($v = 0$). Using $v = u - gt \Rightarrow 0 = 30 - 9.8t \Rightarrow t \approx 3.06 \text{ seconds}$.

When to use: Use this convention in vertical projection calculations to avoid sign errors when solving for time or height.

Memory Tricks — Never Forget These

Trick: Three quick memory triggers to maintain accuracy when solving dynamics problems under pressure.

Steps to Remember:

The SUVAT variable checklist: Before writing any equations, list the five parameters on your exam paper: S (displacement), U (initial velocity), V (final velocity), A (acceleration), and T (time). Identify which three are known and which one is required.
The 'G-Oppose' Rule: Remember: if you throw a ball **Up**, gravity opposes the upward motion, so use **$-g$**. If the ball falls **Down**, gravity assists the downward motion, so use **$+g$**.
Inclined acceleration component: An incline has an angle $\theta$ with the horizontal. Slide down a hill? Gravity is reduced by a factor of the sine of the angle. Acceleration is always $g \sin \theta$.

Mnemonic: Check your SUVAT variables | -g for rising, +g for falling | a = g sinθ on a smooth incline

Common Mistakes

Avoid these common errors to protect your marks in the NEB exam.

Incorrect Unit Conversion: Forgetting to convert velocities given in km/h to m/s. Always multiply km/h by $\frac{5}{18}$ to convert to m/s before calculation.
Confusing t-th second with total time: Confusing the displacement in a specific second $s_t$ with the total displacement $s$ over $t$ seconds. For example, the distance covered in the 5th second is calculated using $s_t$, whereas the distance covered in 5 seconds requires the $s$ equation.
Double accounting for Gravity: Using a negative sign for gravity ($v = u - gt$) and then substituting $g = -9.8 \text{ ms}^{-2}$ again, resulting in a positive value. Use the negative sign in the formula with $g = 9.8 \text{ ms}^{-2}$.
Wrong Initial Velocity for Dropped Objects: Setting $u$ to a non-zero value when an object is 'dropped' or 'released from rest'. For these scenarios, always set $u = 0$.
Misinterpreting Incline Ratios: Misinterpreting 'an incline of 1 in 100' as the angle $\theta = 1^\circ$. This description means $\sin \theta = \frac{1}{100} = 0.01$.

NEB Exam Traps — These Appear Every Year

NEB Exam Warning

Examiners regularly test these specific cases. Keep them in mind during your preparation.

Object Dropped from a Rising Vessel: A stone is dropped from a balloon rising vertically at a constant velocity $u_{\text{b}}$. The dropped stone inherits the balloon's velocity, so its initial velocity is $+u_{\text{b}}$ upward, not zero.
Velocity at Maximum Height: A vertically projected body reaches its peak. Its velocity at this point is $0 \text{ ms}^{-1}$, but its acceleration is still $g = 9.8 \text{ ms}^{-2}$ downwards. Acceleration is never zero at the peak.
Meeting Point of Two Bodies: Problems where one body is dropped from a height while another is projected upwards. Set the sum of their displacements equal to the total height ($h_1 + h_2 = H$) and solve for the time $t$.
Double-Stage Vehicle Motion: A car accelerates from rest to a maximum velocity, travels at constant speed, and then brakes to a stop. Divide this problem into three separate stages, each with its own initial and final conditions.

Question Recognition Guide

Keyword / Phrase in Question	Use This
A vehicle 'starts from rest' or 'slides from the top'	Set initial velocity u = 0
A moving vehicle 'brakes to a stop' or 'comes to rest'	Set final velocity v = 0
A projected body reaches 'maximum height'	Set velocity at peak v = 0, and use acceleration a = −g
An object is 'dropped' or 'falls freely'	Set initial velocity u = 0, and use acceleration a = +g
An incline described as '1 in n'	Set sinθ = 1/n, and use acceleration down the plane as a = g sinθ
Calculate displacement in 'the 5th second' of motion	Use the t-th second formula: s_t = u + 1/2 a(2t - 1) with t = 5
A stone is dropped from a balloon rising at speed v_b	Set the stone's initial velocity u = +v_b (upward direction)
A bullet 'loses 50% of its velocity' after penetrating distance x	Set up a two-part equation using v² = u² + 2as with final velocity v = u/2

Solved Examples (Easy → Hard)

1

A motorcycle accelerates from rest with a uniform acceleration of 4 m/s² for 5 seconds. Find the distance covered during this time.

Easy

Solution

Identify parameters:
$u = 0$ (starts from rest), $a = 4 \text{ ms}^{-2}$, $t = 5 \text{ s}$.

Apply displacement-time relation:
$$s = ut + \frac{1}{2}at^2$$
$$s = (0 \times 5) + \frac{1}{2} \times 4 \times (5)^2$$
$$s = 0 + 2 \times 25 = 50 \text{ m}$$

∴ Distance covered = 50 m

2

A ball is thrown vertically upwards with an initial velocity of 20 m/s. Calculate the time it takes to reach its maximum height. [Take g = 10 m/s²]

Easy

Solution

Identify parameters:
$u = 20 \text{ ms}^{-1}$, $v = 0$ (velocity is zero at peak), $a = -g = -10 \text{ ms}^{-2}$.

Apply velocity-time relation:
$$v = u + at \implies 0 = 20 - 10t$$
$$10t = 20 \implies t = 2 \text{ s}$$

∴ Time to reach max height = 2 seconds

3

A passenger train starts from rest and accelerates uniformly at 2 m/s² for 10 seconds. It then travels at a constant speed for 30 seconds, and is finally brought to rest in 5 seconds by the application of brakes. Calculate the total distance covered.

Medium

Solution

Stage 1 — Acceleration:
$u_1 = 0$, $a_1 = 2 \text{ ms}^{-2}$, $t_1 = 10 \text{ s}$.
$$s_1 = u_1t_1 + \frac{1}{2}a_1t_1^2 = 0 + \frac{1}{2}(2)(10)^2 = 100 \text{ m}$$
Velocity at the end of acceleration:
$$v_1 = u_1 + a_1t_1 = 0 + 2(10) = 20 \text{ ms}^{-1}$$

Stage 2 — Constant Speed:
The train travels at $v = 20 \text{ ms}^{-1}$ for $t_2 = 30 \text{ s}$.
$$s_2 = v \times t_2 = 20 \times 30 = 600 \text{ m}$$

Stage 3 — Deceleration:
$u_3 = 20 \text{ ms}^{-1}$, $v_3 = 0$, $t_3 = 5 \text{ s}$.
Using average velocity:
$$s_3 = \left( \frac{u_3 + v_3}{2} \right) \times t_3 = \left( \frac{20 + 0}{2} \right) \times 5 = 10 \times 5 = 50 \text{ m}$$

Total Distance:
$$s_{\text{total}} = s_1 + s_2 + s_3 = 100 + 600 + 50 = 750 \text{ m}$$

∴ Total distance = 750 m

4

A sliding block starts from rest at the top of a smooth inclined plane of length 8 m and inclination 30°. Find the velocity with which it reaches the bottom of the incline. [Take g = 9.8 m/s²]

Medium

Solution

Step 1 — Calculate acceleration down the incline:
$$a = g \sin \theta = 9.8 \sin 30^\circ = 9.8 \times 0.5 = 4.9 \text{ ms}^{-2}$$

Step 2 — Apply velocity-displacement relation:
$u = 0$, $s = 8$ m, $a = 4.9 \text{ ms}^{-2}$.
$$v^2 = u^2 + 2as \implies v^2 = 0 + 2(4.9)(8)$$
$$v^2 = 9.8 \times 8 = 78.4$$
$$v = \sqrt{78.4} \approx 8.85 \text{ ms}^{-1}$$

∴ Velocity at bottom = 8.85 m/s

5

A stone is dropped from a hot-air balloon that is rising vertically with a velocity of 12 m/s. If the balloon is at a height of 100 m above the ground when the stone is released, calculate the time taken by the stone to hit the ground. [Take g = 10 m/s²]

Hard

Solution

Step 1 — Establish coordinate signs:
Define the vertically upward direction as positive (+y).
• Initial velocity of the stone: $u = +12 \text{ ms}^{-1}$ (inherits upward motion of balloon)
• Displacement to the ground: $s = -100$ m (ends up below release point)
• Acceleration due to gravity: $a = -g = -10 \text{ ms}^{-2}$

Step 2 — Set up displacement equation:
$$s = ut + \frac{1}{2}at^2 \implies -100 = 12t + \frac{1}{2}(-10)t^2$$
$$-100 = 12t - 5t^2 \implies 5t^2 - 12t - 100 = 0$$

Step 3 — Solve the quadratic equation:
Applying the quadratic formula:
$$t = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(5)(-100)}}{2(5)}$$
$$t = \frac{12 \pm \sqrt{144 + 2000}}{10} = \frac{12 \pm \sqrt{2144}}{10}$$
$$t \approx \frac{12 \pm 46.30}{10}$$
Since time $t$ must be positive:
$$t \approx \frac{12 + 46.30}{10} = 5.83 \text{ s}$$

∴ Time to hit the ground = 5.83 seconds

MCQ Practice

Mixed Level — MCQ

1

Which of the following physical quantities is a vector?

Correct!

Incorrect. Correct: Velocity

Solution

Velocity is defined as the time rate of change of displacement, which requires both magnitude and direction, making it a vector.

∴ Velocity

2

A body dropped from a height h reaches the ground in time t. The value of h is given by:

Correct!

Incorrect. Correct: 1/2 gt²

Solution

Using $s = ut + \frac{1}{2}at^2$ with $u = 0$, $s = h$, and $a = g$, we get $h = \frac{1}{2}gt^2$.

∴ 1/2 gt²

3

If a particle moves with a constant velocity, its acceleration must be:

Correct!

Incorrect. Correct: Zero

Solution

Acceleration is the derivative of velocity: $a = \frac{dv}{dt}$. Since $v$ is constant, $a = \frac{d}{dt}(\text{const}) = 0$.

∴ Zero

4

The distance covered in successive seconds (t = 1, 2, 3...) by a body starting from rest with uniform acceleration a forms an arithmetic progression with a common difference of:

Correct!

Incorrect. Correct: a

Solution

The distance in the $t$-th second is $s_t = u + \frac{1}{2}a(2t-1)$. For successive seconds, the difference is $s_{t+1} - s_t = [u + \frac{1}{2}a(2t+1)] - [u + \frac{1}{2}a(2t-1)] = a$.

∴ a

5

The acceleration of a body sliding down a smooth inclined plane of inclination θ is:

Correct!

Incorrect. Correct: g sinθ

Solution

Resolving the weight vector along the incline yields a force of $mg \sin \theta$. Applying Newton's second law ($F = ma$) gives $a = g \sin \theta$.

∴ g sinθ

6

A bullet penetrates a block of wood and loses half of its velocity after traveling 3 cm. How much further will it penetrate before coming to rest?

Correct!

Incorrect. Correct: 1 cm

Solution

First part: $(u/2)^2 = u^2 + 2as \implies 6as = -3u^2/4 \implies a = -u^2/8$ (for $s=3$).
Second part: $0 = (u/2)^2 + 2ax \implies 2\left(-\frac{u^2}{8}\right)x = -\frac{u^2}{4} \implies \frac{u^2}{4}x = \frac{u^2}{4} \implies x = 1 \text{ cm}$.

∴ 1 cm

7

The time of flight of a body projected vertically upwards with velocity u is:

Correct!

Incorrect. Correct: 2u/g

Solution

Time of ascent $t_a = u/g$. In a vacuum, time of descent $t_d = t_a = u/g$. Total time of flight $T = t_a + t_d = 2u/g$.

∴ 2u/g

8

If the displacement of a particle is given by s = 2t³ − 3t², its acceleration at t = 2 s is:

Correct!

Incorrect. Correct: 18 ms⁻²

Solution

$v = \frac{ds}{dt} = 6t^2 - 6t$. $a = \frac{dv}{dt} = 12t - 6$. At $t=2$ s, $a = 12(2) - 6 = 18 \text{ ms}^{-2}$.

∴ 18 ms⁻²

9

A balloon is rising vertically with a velocity of 10 m/s. A stone is dropped from it. The initial velocity of the stone relative to the ground is:

Correct!

Incorrect. Correct: 10 m/s upwards

Solution

An object released from a moving platform inherits the instantaneous velocity of that platform. Thus, the stone has an initial upward velocity of 10 m/s.

∴ 10 m/s upwards

10

When an incline is specified as '1 in 20', it mathematically means that:

Correct!

Incorrect. Correct: sinθ = 1/20

Solution

An incline of '1 in n' means for every $n$ units traveled along the inclined surface, the vertical height increases by 1 unit. This defines $\sin \theta = \frac{1}{n}$.

∴ sinθ = 1/20

Short Questions (2 Marks Each)

1

Define retardation and give a real-life example.

[2 marks]

Solution

Retardation is negative acceleration, occurring when the velocity of a moving body decreases over time.
Example: A driver applying brakes to stop a car at a red traffic light.

∴ Negative acceleration; e.g., a car braking to a stop.

2

Prove that the time of ascent is equal to the time of descent for a body projected vertically upwards in a vacuum.

[2 marks]

Solution

• Time of ascent ($t_a$): using $v = u - gt_a$ with $v=0 \implies t_a = \frac{u}{g}$.
• Time of descent ($t_d$): falls from height $h = \frac{u^2}{2g}$ with initial velocity $0$. Using $h = \frac{1}{2}gt_d^2 \implies \frac{u^2}{2g} = \frac{1}{2}gt_d^2 \implies t_d = \frac{u}{g}$.
Thus, $t_a = t_d = \frac{u}{g}$.

∴ Both times equal u/g, proving they are equal.

3

What is meant by a 'particle' in mechanics?

[2 marks]

Solution

In mechanics, a particle is an idealized body of negligible dimensions. It is treated as a geometrical point with mass, where its size is small compared to the distance it travels.

∴ An idealized mass point of negligible size relative to the distance of its motion.

4

Distinguish between speed and velocity.

[2 marks]

Solution

• Speed: A scalar quantity representing the rate at which an object covers distance, independent of direction.
• Velocity: A vector quantity representing the rate of change of displacement, which includes both speed and direction.

∴ Speed is scalar (distance/time); velocity is vector (displacement/time).

5

A stone is dropped into a well. Why is the sound of the splash heard slightly after the stone hits the water?

[2 marks]

Solution

When the stone hits the water, the sound of the splash is produced instantly. However, this sound must travel up the depth of the well to reach the observer's ears. This travel time is $t = \frac{\text{depth}}{v_{\text{sound}}}$, causing a slight delay.

∴ Sound takes time to travel from the bottom of the well to the observer's ears.

6

What is the acceleration of a vertically projected body at its highest point?

[2 marks]

Solution

At the highest point, the instantaneous velocity of the body is $0 \text{ ms}^{-1}$. However, the acceleration is not zero; it is still $g = 9.8 \text{ ms}^{-2}$ directed vertically downwards, due to Earth's continuous gravitational pull.

∴ Acceleration is 9.8 ms⁻² downwards.

7

Define uniform acceleration and state its SI unit.

[2 marks]

Solution

Uniform acceleration occurs when the velocity of a body changes by equal amounts in equal intervals of time, no matter how small the intervals are.
Its SI unit is **meter per second squared ($\text{m/s}^2$ or $\text{ms}^{-2}$)**.

∴ Equal change in velocity in equal intervals of time; unit is ms⁻².

8

Show that if the inclination of a smooth inclined plane becomes 90°, the motion reduces to free fall.

[2 marks]

Solution

The acceleration of a body sliding down a smooth inclined plane is $a = g \sin \theta$.
When the inclination $\theta = 90^\circ$ (vertical plane):
$$a = g \sin 90^\circ = g(1) = g$$
This is the acceleration of a free-falling body, proving the relation.

∴ At θ = 90°, a = g sin90° = g, which is free fall.

9

Write down the expression for the distance covered in the n-th second and state the physical dimensions of the constant terms.

[2 marks]

Solution

The formula is $s_n = u + \frac{1}{2}a(2n-1)$.
Although this expression appears to add quantities with different dimensions, the constant '1' actually has the dimension of time ($[T]$), and the equation represents a rate of displacement per unit time, maintaining dimensional consistency.

∴ sn = u + 1/2 a(2n-1); dimensionally consistent when accounting for unit time.

10

Does the mass of a body affect its acceleration during free fall in a vacuum? Explain.

[2 marks]

Solution

No, the mass does not affect the acceleration. In a vacuum, there is no air resistance. Since $a = g$ for all falling bodies regardless of mass, a heavy object and a light object will fall with the exact same acceleration.

∴ No, in a vacuum all bodies fall with the same gravitational acceleration g, independent of mass.

Long Questions (4–5 Marks Each)

1

Derive the equations of motion (i) v = u + at, (ii) s = ut + 1/2 at², and (iii) v² = u² + 2as using the calculus (integration) method.

[5 marks]

Solution

See Section 6 for the detailed mathematical derivations of these three equations using definite integration with boundary conditions.

∴ Derivations completed using calculus method [Proved]

2

Derive an expression for the distance traveled by a uniformly accelerating body in its t-th second.

[4 marks]

Solution

See Section 6 for the detailed step-by-step mathematical derivation of $s_t = u + \frac{1}{2}a(2t-1)$ by subtracting $s(t-1)$ from $s(t)$.

∴ st = u + 1/2 a(2t − 1) [Proved]

3

A stone is dropped from a hot-air balloon rising vertically with a velocity of 12 m/s at a height of 100 m above the ground. Calculate the time taken by the stone to hit the ground. [Take g = 10 m/s²]

[5 marks]

Solution

See Example 5 in this section for the complete step-by-step algebraic solution of this vertical projection problem.

∴ Time taken = 5.83 seconds

4

A car covers 20 m in the 3rd second and 30 m in the 5th second of its motion under uniform acceleration. Calculate its initial velocity and acceleration.

[4 marks]

Solution

Step 1 — Set up equations using the t-th second formula:
$$s_t = u + \frac{1}{2}a(2t - 1)$$
• For $t = 3$ s, $s_3 = 20$ m:
$$20 = u + \frac{1}{2}a(2(3) - 1) \implies 20 = u + 2.5a \quad \text{--- (i)}$$
• For $t = 5$ s, $s_5 = 30$ m:
$$30 = u + \frac{1}{2}a(2(5) - 1) \implies 30 = u + 4.5a \quad \text{--- (ii)}$$

Step 2 — Solve for acceleration (a):
Subtract equation (i) from (ii):
$$10 = 2a \implies a = 5 \text{ ms}^{-2}$$

Step 3 — Solve for initial velocity (u):
Substitute $a = 5$ into (i):
$$20 = u + 2.5(5) \implies 20 = u + 12.5 \implies u = 7.5 \text{ ms}^{-1}$$

∴ Initial velocity = 7.5 m/s; Acceleration = 5 m/s²

5

A body is projected vertically upwards with a velocity of 49 m/s. Find (i) the maximum height reached, (ii) the time taken to reach the maximum height, and (iii) the velocity with which it strikes the ground. [Take g = 9.8 m/s²]

[5 marks]

Solution

Step 1 — (i) Calculate maximum height (h):
$u = 49 \text{ ms}^{-1}$, $v = 0$ (at peak), $a = -g = -9.8 \text{ ms}^{-2}$.
$$v^2 = u^2 + 2as \implies 0 = 49^2 - 2(9.8)h$$
$$19.6h = 2401 \implies h = \frac{2401}{19.6} = 122.5 \text{ m}$$

Step 2 — (ii) Calculate time of ascent (t):
$$v = u + at \implies 0 = 49 - 9.8t \implies 9.8t = 49 \implies t = 5 \text{ s}$$

Step 3 — (iii) Find striking velocity:
In a vacuum, the motion is symmetrical. The time of descent equals the time of ascent ($5$ s), and the body strikes the ground with the same speed it was projected with, but in the opposite direction:
$$v_{\text{strike}} = -49 \text{ ms}^{-1} \text{ (downwards)}$$

∴ Max Height = 122.5 m; Time = 5 s; Striking Velocity = −49 m/s

6

A bullet loses half of its velocity after penetrating 3 cm into a uniform wooden block. Find how much further it will penetrate before coming to rest, assuming uniform retardation.

[4 marks]

Solution

Step 1 — Set up the equation for the first stage:
Let $u$ be the initial velocity. The velocity after $s_1 = 3$ cm is $v_1 = \frac{u}{2}$.
$$v_1^2 = u^2 + 2as_1 \implies \left( \frac{u}{2} \right)^2 = u^2 + 2a(3)$$
$$\frac{u^2}{4} = u^2 + 6a \implies 6a = -\frac{3u^2}{4} \implies a = -\frac{u^2}{8} \quad \text{--- (i)}$$

Step 2 — Set up the equation for the second stage:
Let the bullet travel an additional distance $s_2$ before stopping ($v_2 = 0$).
$$v_2^2 = v_1^2 + 2as_2 \implies 0 = \left( \frac{u}{2} \right)^2 + 2as_2 \implies 0 = \frac{u^2}{4} + 2as_2$$
Substitute $a = -\frac{u^2}{8}$ into the equation:
$$0 = \frac{u^2}{4} + 2\left( -\frac{u^2}{8} \right)s_2 \implies 0 = \frac{u^2}{4} - \frac{u^2}{4} s_2$$
$$\frac{u^2}{4} s_2 = \frac{u^2}{4} \implies s_2 = 1 \text{ cm}$$

∴ Further penetration = 1 cm

7

A body slides down a smooth inclined plane of length l and inclination θ, starting from rest at the top. Derive expressions for (i) its acceleration, (ii) its velocity at the bottom, and (iii) the time taken to reach the bottom.

[5 marks]

Solution

Step 1 — (i) Acceleration:
Resolving gravity parallel to the plane: $a = g \sin \theta$.

Step 2 — (ii) Velocity at the bottom (v):
$u = 0$, $s = l$, $a = g \sin \theta$.
$$v^2 = u^2 + 2as \implies v^2 = 0 + 2(g \sin \theta)l \implies v = \sqrt{2gl \sin \theta}$$

Step 3 — (iii) Time to reach the bottom (t):
$$s = ut + \frac{1}{2}at^2 \implies l = 0 + \frac{1}{2}(g \sin \theta)t^2$$
$$t^2 = \frac{2l}{g \sin \theta} \implies t = \sqrt{\frac{2l}{g \sin \theta}}$$.

∴ a = g sinθ; v = √(2gl sinθ); t = √(2l / g sinθ)

8

Two stones are positioned vertically: one is dropped from a tower of height 150 m, and at the same instant, another is projected vertically upwards from the foot of the tower with a velocity of 50 m/s. Find when and where they will meet. [Take g = 10 m/s²]

[5 marks]

Solution

Step 1 — Formulate displacement equations:
Let the stones meet at time $t$ at a height $h$ from the ground.
• Stone 1 (Dropped from top): travels distance $y_1$ downwards.
$$y_1 = \frac{1}{2}gt^2 = \frac{1}{2}(10)t^2 = 5t^2$$
• Stone 2 (Projected from bottom): travels height $y_2$ upwards.
$$y_2 = ut - \frac{1}{2}gt^2 = 50t - 5t^2$$

Step 2 — Solve for time (t):
The sum of their vertical travels must equal the height of the tower:
$$y_1 + y_2 = 150 \implies 5t^2 + (50t - 5t^2) = 150$$
$$50t = 150 \implies t = 3 \text{ s}$$

Step 3 — Calculate meeting height:
$$y_2 = 50(3) - 5(3)^2 = 150 - 45 = 105 \text{ m from the ground}$$

∴ They meet after 3 seconds at a height of 105 m from the ground

9

Prove that for a particle moving with uniform acceleration, if the distances covered in the p-th, q-th, and r-th seconds are a, b, and c respectively, then: a(q − r) + b(r − p) + c(p − q) = 0.

[5 marks]

Solution

Step 1 — Express distances using the t-th second formula:
Let $u$ be the initial velocity and $f$ be the uniform acceleration.
• $a = u + \frac{1}{2}f(2p - 1)$ ... (i)
• $b = u + \frac{1}{2}f(2q - 1)$ ... (ii)
• $c = u + \frac{1}{2}f(2r - 1)$ ... (iii)

Step 2 — Find differences between equations:
• Subtract (ii) from (i):
$$a - b = \frac{1}{2}f[(2p - 1) - (2q - 1)] = f(p - q) \implies p - q = \frac{a - b}{f}$$ ... (iv)
• Subtract (iii) from (ii):
$$b - c = f(q - r) \implies q - r = \frac{b - c}{f}$$ ... (v)
• Subtract (i) from (iii):
$$c - a = f(r - p) \implies r - p = \frac{c - a}{f}$$ ... (vi)

Step 3 — Substitute into target expression:
$$a(q - r) + b(r - p) + c(p - q) = a\left( \frac{b - c}{f} \right) + b\left( \frac{c - a}{f} \right) + c\left( \frac{a - b}{f} \right)$$
$$= \frac{1}{f} [a(b - c) + b(c - a) + c(a - b)]$$
$$= \frac{1}{f} [ab - ac + bc - ab + ac - bc] = \frac{1}{f} [0] = 0 \quad \text{[Proved]}$$.

∴ Algebraic substitution yields 0 [Proved]

10

Explain the difference between Kinematics and Kinetics, using real-life physical examples to illustrate both sub-branches of dynamics.

[4 marks]

Solution

• Kinematics: Describes motion without considering the forces causing it. It focuses on variables like displacement, velocity, acceleration, and time.
Example: Calculating that a car traveling at 20 m/s with a deceleration of 2 m/s² takes 10 seconds to stop.
• Kinetics: Studies the forces that cause or change motion, linking force and mass to acceleration.
Example: Calculating the friction force exerted by the brakes and road surface to decelerate a 1500 kg car at 2 m/s².

∴ Kinematics describes motion variables; kinetics analyzes the forces causing motion.

Past NEB Board Questions

1

A stone is dropped from a height of 80 m. Find the time taken by it to reach the ground and its velocity just before hitting the ground. [Take g = 10 m/s²]

NEB 2079 [4 marks]

Solution

Step 1 — Calculate time (t):
$u = 0$, $s = 80$ m, $a = g = 10 \text{ ms}^{-2}$.
$$s = ut + \frac{1}{2}at^2 \implies 80 = 0 + \frac{1}{2}(10)t^2$$
$$80 = 5t^2 \implies t^2 = 16 \implies t = 4 \text{ s}$$

Step 2 — Calculate velocity (v):
$$v = u + at \implies v = 0 + 10(4) = 40 \text{ ms}^{-1}$$.

∴ Time = 4 seconds; Velocity = 40 m/s

2

A body projected vertically upwards from the ground passes a point at a height of 40 m twice with an interval of 4 seconds. Calculate its initial velocity. [Take g = 10 m/s²]

NEB 2078 [4 marks]

Solution

Step 1 — Relate the time interval to the trajectory:
Let $t_1$ and $t_2$ be the two times the body passes the height $h = 40$ m. The time interval is $t_2 - t_1 = 4$ seconds.
Using $s = ut - \frac{1}{2}gt^2$ with $s = 40$:
$$40 = ut - 5t^2 \implies 5t^2 - ut + 40 = 0$$
This quadratic equation has roots $t_1$ and $t_2$.

Step 2 — Apply properties of quadratic roots:
• Sum of roots: $t_1 + t_2 = \frac{u}{5}$
• Product of roots: $t_1 t_2 = \frac{40}{5} = 8$

Step 3 — Solve for u:
Using the identity $(t_2 - t_1)^2 = (t_1 + t_2)^2 - 4t_1 t_2$:
$$4^2 = \left( \frac{u}{5} \right)^2 - 4(8)$$
$$16 = \frac{u^2}{25} - 32 \implies 48 = \frac{u^2}{25}$$
$$u^2 = 48 \times 25 = 1200 \implies u = \sqrt{1200} \approx 34.64 \text{ ms}^{-1}$$.

∴ Initial Velocity = 34.64 m/s

3

State the laws of motion on a smooth inclined plane. A body slides down a smooth inclined plane of length 10 m and inclination 30° starting from rest. Find the time taken to reach the bottom. [Take g = 9.8 m/s²]

NEB 2077 [4 marks]

Solution

Step 1 — Calculate acceleration:
$$a = g \sin \theta = 9.8 \sin 30^\circ = 4.9 \text{ ms}^{-2}$$

Step 2 — Apply displacement equation:
$u = 0$, $s = 10$ m, $a = 4.9 \text{ ms}^{-2}$.
$$s = ut + \frac{1}{2}at^2 \implies 10 = 0 + \frac{1}{2}(4.9)t^2$$
$$10 = 2.45t^2 \implies t^2 = \frac{10}{2.45} \approx 4.08$$
$$t = \sqrt{4.08} \approx 2.02 \text{ s}$$.

∴ Time taken = 2.02 seconds

4

Prove that v² = u² + 2as for a body moving with uniform acceleration, using both graphical and analytical (calculus) methods.

NEB 2076 [4 marks]

Solution

• Calculus Method: See Section 6 for the derivation using $a = v \frac{dv}{ds}$.
• Graphical Method: In a velocity-time graph with constant acceleration, displacement $s$ is the area of the trapezium under the line:
$s = \frac{u+v}{2} \times t$. Substitute $t = \frac{v-u}{a}$ from the definition of acceleration:
$s = \left( \frac{v+u}{2} \right) \left( \frac{v-u}{a} \right) = \frac{v^2 - u^2}{2a} \implies v^2 = u^2 + 2as$.

∴ Analytical and graphical derivations completed [Proved]

Chapter Test — Full Mixed Paper

Attempt this test in 45 minutes without looking at your notes. This matches the standard NEB exam format.

1 Section A — MCQs (1 Mark Each)

Mixed Level — MCQ

1

The SI unit of acceleration is:

Correct!

Incorrect. Correct: ms⁻²

Solution

Acceleration is the rate of change of velocity per unit time: $\frac{\text{m/s}}{\text{s}} = \text{m/s}^2$ or $\text{ms}^{-2}$.

∴ ms⁻²

2

An incline of '3 in 5' means that:

Correct!

Incorrect. Correct: sinθ = 3/5

Solution

An incline of $y$ in $x$ is defined as $\sin \theta = \frac{y}{x}$, so '3 in 5' means $\sin \theta = 3/5 = 0.6$.

∴ sinθ = 3/5

3

At maximum height, the velocity of a vertically projected body is:

Correct!

Incorrect. Correct: Zero

Solution

At maximum height, the body momentarily stops to change direction, meaning its velocity is $0 \text{ ms}^{-1}$.

∴ Zero

2 Section B — Short Questions (2 Marks Each)

1

Define dynamics and list its two sub-branches.

[2 marks]

Solution

Dynamics is the branch of mechanics that studies the motion of physical bodies under the action of forces.
It is divided into two sub-branches:
1. Kinematics: Motion without considering forces.
2. Kinetics: Motion considering the forces causing it.

∴ Studies motion with forces; divided into kinematics and kinetics.

2

What is the sign of acceleration due to gravity for a vertically rising body?

[2 marks]

Solution

For a vertically rising body, the direction of motion is upwards, while gravity acts downwards. Since gravity opposes the motion, the acceleration is negative relative to the direction of projection: $a = -g$.

∴ Negative (−g) relative to the upward projection.

3 Section C — Long Questions (4–5 Marks Each)

1

Derive the displacement-time relation s = ut + 1/2 at² using the integration method.

[4 marks]

Solution

See Section 6 for the detailed derivation starting with $ds = (u+at)dt$ and integrating within limits.

∴ s = ut + 1/2 at² derived [Proved]

2

A ball dropped from a height of 80 m takes how long to hit the ground? [Take g = 10 m/s²]

[4 marks]

Solution

$u = 0$, $s = 80$ m, $a = g = 10 \text{ ms}^{-2}$.
$$s = ut + \frac{1}{2}at^2 \implies 80 = 0 + \frac{1}{2}(10)t^2$$
$$80 = 5t^2 \implies t^2 = 16 \implies t = 4 \text{ s}$$.

∴ Time taken = 4 seconds

One-Page Revision Cheat Sheet

1

Kinematic Relations: $v = u + at$, $s = ut + \frac{1}{2}at^2$, $v^2 = u^2 + 2as$ — only valid for constant acceleration.

2

t-th Second displacement: $s_t = u + \frac{1}{2}a(2t - 1)$ — gives displacement during a single specific second.

3

Rising gravity: Use $a = -g$ when an object is projected upwards.

4

Falling gravity: Use $a = +g$ when an object is falling downwards.

5

Peak velocity: $v = 0$ at maximum height; acceleration is still $g$ downwards.

6

Smooth Incline: Acceleration down the plane is $a = g \sin \theta$.

7

Inclined components: $g \sin \theta$ acts down the incline; $g \cos \theta$ is balanced by the normal reaction.

8

Release from rest: Always set $u = 0$ for dropped objects.

9

Incline ratio: '1 in n' incline means $\sin \theta = 1/n$.

10

Unit Check: Convert km/h to m/s immediately by multiplying by 5/18.