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Chapter 2 of 7

Kinematics and Kinetics

In the RRB JE Civil CBT-2 syllabus under Engineering Mechanics · 2 parts

Kinematics of Particles · Kinetics of Particles — Newton's Laws & D'Alembert's Principle

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📑 Contents (19 sections)

Part 1 of 2

Kinematics of Particles

Last reviewed 16 Sept 2026 · 5 min read

Kinematics

Kinematics describes motion — position, velocity and acceleration — without asking what forces cause it (that is kinetics). A particle is a body whose size does not matter for the motion being studied.

DefinitionBasic quantities

Distance — total path length (scalar). Displacement — change in position (vector).
Speed — rate of change of distance. Velocity $v = \frac{d x}{d t}$ (vector).
Acceleration $a = \frac{d v}{d t} = \frac{d ^{2} x}{d t ^{2}} = v \frac{d v}{d x}$ .
Average velocity = total displacement ÷ total time.

Rectilinear motion with uniform acceleration

FormulaEquations of motion (constant

a

)

v = u + a t

s = u t + \frac{1}{2} a t^{2}

v^{2} = u^{2} + 2 a s

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

Distance covered in the $n$ th second: $s_{n} = u + \frac{a}{2} (2 n - 1)$

Motion under gravity

Take $g$ = 9.81 m/s² downward (neglecting air resistance).

Body thrown up with speed $u$ : time to reach the top $= u / g$ ; maximum height $= u^{2} /2 g$ ; total time of flight (back to the start) $= 2 u / g$ ; it returns with speed $u$ .
Body dropped from height $h$ : time $= 2 h / g$ ; striking velocity $= 2 g h$ .

Variable acceleration

When $a$ changes, integrate:

$a = f (t)$ : $v = \int a d t$ , $x = \int v d t$ .
$a = f (v)$ : $d t = \frac{d v}{f ( v )}$ or $d x = \frac{v d v}{f ( v )}$ .
$a = f (x)$ : $v d v = f (x) d x$ .

Motion graphs

Graph	Slope gives	Area gives
Displacement–time	Velocity	—
Velocity–time	Acceleration	Displacement
Acceleration–time	Rate of change of acceleration (jerk)	Change in velocity

Relative velocity

The velocity of A relative to B:

v_{A / B} = v_{A} - v_{B}

Classic problems: two trains, rain and umbrella angle, boat crossing a river.

Boat crossing a river of width $d$ , boat speed $v_{b}$ in still water, river speed $v_{r}$ :

Shortest time: head straight across; $t = d / v_{b}$ ; drift downstream $= v_{r} d / v_{b}$ .
Shortest path (straight across, $v_{b} > v_{r}$ ): head upstream at angle $θ$ with $sin θ = v_{r} / v_{b}$ to the perpendicular; $t = \frac{d}{v _{b}^{2} - v _{r}^{2}}$ .

Part 2 of 2

Kinetics of Particles — Newton's Laws & D'Alembert's Principle

Last reviewed 16 Sept 2026 · 5 min read

Newton's laws of motion

First law (inertia): a body stays at rest or in uniform motion in a straight line unless acted on by an unbalanced external force.
Second law: the rate of change of momentum is proportional to the applied force and takes place in its direction: $F = \frac{d ( m v )}{d t} = ma$ for constant mass.
Third law: to every action there is an equal and opposite reaction (acting on the other body).

Units: 1 newton (N) is the force that gives 1 kg an acceleration of 1 m/s². Weight $W = m g$ ; mass is constant, weight changes with $g$ .

Equation of motion

\sum F = m a

Draw the free-body diagram, choose the direction of motion as positive, include every force (applied loads, weight, normal reaction, friction, tension), then apply $\sum F = ma$ along the direction of motion and $\sum F = 0$ perpendicular to it.

D'Alembert's principle

Rewrite $\sum F = ma$ as $\sum F - ma = 0$ . The term $- ma$ is an imaginary inertia force acting opposite to the acceleration.

FormulaD'Alembert's principle

A body in accelerated motion can be treated as if it were in equilibrium under the real forces plus an inertia force $m a$ applied opposite to the acceleration:

\sum F + (- ma) = 0

This turns dynamics problems into statics problems (dynamic equilibrium) — useful for connected bodies and rotating systems.

Lift (elevator) problems

A person of mass $m$ stands on a weighing machine in a lift. The reading equals the normal reaction $R$ :

Motion of lift	Reaction (apparent weight)
At rest or uniform velocity	$R = m g$
Accelerating upward (or decelerating while moving down)	$R = m (g + a)$
Accelerating downward (or decelerating while moving up)	$R = m (g - a)$
Free fall ( $a = g$ )	$R = 0$ (weightlessness)

Cable tension for a lift of mass $M$ follows the same pattern: $T = M (g \pm a)$ .

Connected bodies

Two masses over a smooth pulley (Atwood machine)

Masses $m_{1} > m_{2}$ hang over a frictionless, massless pulley:

a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}, T = \frac{2 m _{1} m _{2} g}{m _{1} + m _{2}}

The pulley support carries $2 T$ .

Mass on a table pulling a hanging mass

Mass $m_{1}$ on a horizontal table with friction coefficient $μ$ , connected over a pulley to $m_{2}$ hanging:

a = \frac{( m _{2} - μ m _{1} ) g}{m _{1} + m _{2}}, T = m_{1} (a + μg)

Mass on an inclined plane connected to a hanging mass

With $m_{1}$ on a plane at $α$ (moving up the plane), $m_{2}$ hanging:

a = \frac{[ m _{2} - m _{1} ( sin α + μ cos α ) ] g}{m _{1} + m _{2}}

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