Kinematics: Motion in a Straight Line

Kinematics is the branch of mechanics describing motion without considering forces. In straight-line motion, we analyze movement along a single axis.

Key Definitions:

• Displacement (s): Vector quantity measuring change in position from starting point, measured in metres (m). Can be positive or negative.
• Distance: Scalar quantity measuring total path length travelled, always positive.
• Velocity (v): Rate of change of displacement; vector with magnitude and direction (m s⁻¹).
• Speed: Scalar quantity, magnitude of velocity only.
• Acceleration (a): Rate of change of velocity (m s⁻²); positive = speeding up in positive direction, negative = slowing down or speeding up in negative direction.

SUVAT Equations (constant acceleration only):

1. v = u + at (velocity-time relationship)
2. s = ut + ½at² (displacement without final velocity)
3. s = vt - ½at² (displacement without initial velocity)
4. v² = u² + 2as (relates velocity and displacement, no time)
5. s = ½(u + v)t (average velocity method)

Where: s = displacement, u = initial velocity, v = final velocity, a = acceleration, t = time

Mnemonic: "Super Useful Velocity And Time" helps remember the variables.

Cambridge Definition Note: Always state that these equations apply ONLY to uniform (constant) acceleration.

Graphical Representations:

• Displacement-time graphs: gradient = velocity; curved line = acceleration
• Velocity-time graphs: gradient = acceleration; area under curve = displacement
• Acceleration-time graphs: area under curve = change in velocity

Understanding kinematics transforms how we perceive everyday motion. Consider a car accelerating from traffic lights: initially at rest (u = 0), it reaches 15 m s⁻¹ in 6 seconds with constant acceleration. Using v = u + at gives a = 2.5 m s⁻².

Real-World Applications:

1. Emergency Braking Systems: Modern cars calculate stopping distances using v² = u² + 2as. At 30 m s⁻¹ with maximum deceleration -8 m s⁻², the stopping distance is 56.25 m. Anti-lock braking systems optimize this deceleration.

2. Athletics - 100m Sprint: Usain Bolt's world record demonstrates variable acceleration. Elite sprinters accelerate maximally for ~60m, then maintain velocity. Analyzing displacement-time graphs reveals peak velocity occurs before the finish line.

3. Vertical Motion - Parachuting: A skydiver experiences free fall (a = -9.81 m s⁻²) until terminal velocity. The negative acceleration reflects downward direction. Using s = ut + ½at², we calculate distance fallen in first 3 seconds: s = 0 + ½(-9.81)(3²) = -44.1 m (negative = downward).

Helpful Analogy: Think of acceleration as a "velocity changer". Just as a dimmer switch gradually changes light intensity, acceleration gradually changes velocity. Zero acceleration = constant velocity (like a cruise control car).

Sign Convention Critical: Always establish a positive direction at the start. For vertical motion, typically upward = positive, downward = negative. This makes g = -9.81 m s⁻² when considering upward as positive.

Real-world Complication: These equations assume negligible air resistance - acceptable for short times/distances but significant for skydivers or long-range projectiles.

Example 1: Cambridge-Style Acceleration Problem

Question: A train travels from station A, accelerating uniformly from rest. After 40 s, it reaches 25 m s⁻¹. Calculate: (a) acceleration, (b) distance travelled. [4 marks]

Solution: (a) Given: u = 0 m s⁻¹, v = 25 m s⁻¹, t = 40 s, a = ? Using v = u + at 25 = 0 + a(40) a = 25/40 = 0.625 m s⁻² ✓

Examiner note: State the equation used and show substitution clearly for method marks.

(b) Using s = ½(u + v)t (most efficient here) s = ½(0 + 25)(40) = 500 m ✓

Alternative: s = ut + ½at² = 0 + ½(0.625)(40²) = 500 m ✓

Example 2: Braking Problem

Question: A car travelling at 18 m s⁻¹ brakes with uniform deceleration, stopping in 30 m. Calculate the deceleration. [3 marks]

Solution: Given: u = 18 m s⁻¹, v = 0 m s⁻¹, s = 30 m, a = ?

Using v² = u² + 2as (no time given) 0² = 18² + 2a(30) 0 = 324 + 60a 60a = -324 a = -5.4 m s⁻² ✓

Negative indicates deceleration opposite to motion direction.

Example 3: Two-stage Motion

Question: A cyclist accelerates at 2 m s⁻² for 5 s from rest, then travels at constant velocity for 8 s. Find total displacement. [4 marks]

Solution: Stage 1 (acceleration): u = 0, a = 2 m s⁻², t = 5 s s₁ = ut + ½at² = 0 + ½(2)(5²) = 25 m ✓ v = u + at = 0 + 2(5) = 10 m s⁻¹

Stage 2 (constant velocity): v = 10 m s⁻¹, t = 8 s, a = 0 s₂ = vt = 10(8) = 80 m ✓

Total displacement = 25 + 80 = 105 m ✓