8.1 Units, Measurement & Motion

Why Units and Motion Matter for RRB Group D

Physics begins with measurement. Railway recruitment exams frequently ask about the SI units of physical quantities, the difference between scalar and vector quantities, and the equations that govern moving bodies — from a train pulling out of a platform to a ball thrown vertically upward. Roughly 6–8 of the 25 General Science questions touch physics, and a large share of those test units, motion, and basic kinematics.

The SI System of Units

The Système International (SI) defines seven base units from which all other units are derived.

Derived units combine base units. Examples: velocity = m/s, acceleration = m/s², force = kg·m/s² (newton, N), energy = kg·m²/s² (joule, J), power = J/s (watt, W).

Common Prefixes

A common RRB trap is confusing capital vs lowercase symbols: m is milli and M is mega; s is second while S is siemens. Use the correct case.

Scalar and Vector Quantities

• Scalar quantities have magnitude only — distance, speed, mass, time, temperature, work, energy.
• Vector quantities have both magnitude and direction — displacement, velocity, acceleration, force, weight, momentum.

A car travelling 5 km north then 5 km south covers a distance of 10 km but has displacement of zero because it returns to the starting point.

Distance, Displacement, Speed, Velocity, Acceleration

• Distance (m): total path length; scalar; always ≥ displacement.
• Displacement (m): change in position from initial to final point; vector; can be zero or negative.
• Speed (m/s): distance ÷ time; scalar.
• Velocity (m/s): displacement ÷ time; vector.
• Acceleration (m/s²): rate of change of velocity; a = (v − u) / t.

Worked Example

A train starts from rest at a station and reaches 20 m/s in 40 s with uniform acceleration. Find acceleration and distance covered.

Given: u = 0, v = 20 m/s, t = 40 s.

a = (v − u) / t = (20 − 0) / 40 = 0.5 m/s².

Using s = ut + ½at² = 0 + ½ × 0.5 × 40² = ½ × 0.5 × 1600 = 400 m.

Equations of Uniformly Accelerated Motion

For motion with constant acceleration a:

1. v = u + at
2. s = ut + ½at²
3. v² = u² + 2as

where u = initial velocity, v = final velocity, a = acceleration, s = displacement, t = time.

Worked Example: Free Vertical Motion

A ball is thrown vertically upward at 19.6 m/s. Take g = 9.8 m/s² downward. Find the maximum height reached.

At top, final velocity v = 0, acceleration a = −g = −9.8 m/s².

Using v² = u² + 2as: 0 = (19.6)² + 2(−9.8)s → s = (19.6)² / (2 × 9.8) = 384.16 / 19.6 = 19.6 m.

The ball takes t = u / g = 19.6 / 9.8 = 2 s to reach the top and another 2 s to return — total 4 s for the round trip.

Graphical Representation of Motion

Distance-Time Graph

• Uniform motion: straight line through origin; slope = speed.
• Rest: horizontal line (slope = 0).
• Accelerated motion: curved line (parabola) opening upward.

Velocity-Time Graph

• Uniform acceleration: straight line sloping upward; slope = acceleration.
• Uniform velocity: horizontal line; area under = displacement.
• Area under a v-t graph always equals displacement.

A common exam trap: the slope of a distance-time graph gives speed (a scalar), while the slope of a velocity-time graph gives acceleration. Mixing these up is one of the most frequent errors.

Circular Motion

When a body moves in a circular path at constant speed, its velocity is not uniform because direction changes continuously — so it is accelerated motion. The acceleration, directed toward the centre, is called centripetal acceleration, a = v² / r.

Uniform vs Non-Uniform Motion

Uniform motion means equal distances covered in equal intervals of time — speed is constant and acceleration is zero. Non-uniform motion means unequal distances in equal intervals — speed changes, so there is acceleration (which may be positive or negative). A train pulling away from a station is in non-uniform (accelerated) motion; once it settles at its cruising speed on a straight, level track with the throttle held steady, it is approximately in uniform motion. RRB often phrases a question as "a body covers 10 m, 20 m, 30 m in successive equal intervals of 2 s" — that is uniformly accelerated motion, not uniform motion.

Worked Example: Braking (Deceleration)

A locomotive running at 36 km/h applies the brakes and comes to rest in 10 s. Find the deceleration and the distance covered before stopping.

First convert the speed to SI units: 36 km/h = 36 × (1000 m / 3600 s) = 10 m/s.

Given u = 10 m/s, v = 0, t = 10 s.

Deceleration a = (v − u) / t = (0 − 10) / 10 = −1 m/s².

Distance s = ut + ½at² = 10 × 10 + ½ × (−1) × 10² = 100 − 50 = 50 m.

This example shows two units skills the exam loves: converting km/h to m/s (multiply by 5/18) and reading a negative acceleration as deceleration.

Common Misconceptions

• "Zero displacement means zero distance." Wrong. A body can travel a long path and return to its start — distance > 0 but displacement = 0, as in one lap of a circular track.
• "Constant speed means constant velocity." Wrong. Constant speed with changing direction (circular motion) means velocity is changing, so the body is accelerating.
• "Negative acceleration always means slowing down." Not always — it depends on the chosen sign of direction. If a body moving in the negative direction speeds up, its acceleration is also negative but the body is speeding up. The safer statement is: if velocity and acceleration have opposite signs, the body slows down; if they have the same sign, it speeds up.
• "Heavier bodies fall faster." Wrong in the absence of air resistance. All bodies in free fall accelerate at g, regardless of mass.