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Chapter 02 · Part I · Mechanics
Revised on ____________________
Chapter 02 · Kinematics in One Dimension

Motion, its graphs, and the five equations

2 sheets
2 diagrams
32 results

Formulas and conditions only — no derivations, no solved numbers. The diagrams are the reader's own.

I

Definitions

AVERAGE IS NOT INSTANTANEOUS
I.Average velocity  ⟨v⟩ = ΔxΔt displacement ÷ time
II.Average speed  ⟨s⟩ = distanceΔt never negative
III.Instantaneous  v = dxdt , a = dvdt
IV.The third form  a = v dvdx use when a = a(x)
V.Displacement  Δx = ∫ v dt signed area under v–t
VI.Two-leg average  ⟨v⟩ = 2v1v2v1 + v2 equal distances, not equal times
Slope going down the ladder, area coming back up: x → v → a by differentiating, a → v → x by integrating.
Position x(t)Velocity v(t)Acceleration a(t)slopeslopeareaarea↓ differentiateintegrate ↑
II

Constant acceleration

FIVE EQUATIONS, ONE MISSING VARIABLE EACH
I.No x  v = u + a t
II.No v  x = u t + 12 a t²
III.No t  v² = u² + 2 a x
IV.No a  x = u + v2 t
V.No u  x = v t − 12 a t²
VI.n-th second  sn = u + a2 (2n − 1) a distance, not a rate
All five assume a constant. The moment acceleration varies, go back to calculus — no exception, no shortcut.
III

Reading the graphs

MOST MARKS IN THIS CHAPTER
I.x–t slope  = v steep = fast, flat = at rest
II.x–t curvature  concave up ⇒ a > 0
III.v–t slope  = a
IV.v–t area  = Δx below the axis subtracts
V.a–t area  = Δv
VI.Turning point  v = 0 with a ≠ 0 top of a throw
VII.Straight v–t  uniform acceleration
VIII.Discontinuity  a jump in v means an infinite a — a collision, not free motion
tvArea = displacement Δxa = 0slope +aslope −a
IV

Vertical motion

SYMMETRY IS THE SHORTCUT
I.Take up positive  a = − g throughout including at the top
II.Time up  t↑ = ug = t↓
III.Maximum height  H = u²2g
IV.Total flight  T = 2ug same level
V.Speed on return  |v| = u same point, same speed
VI.From a height h  h = − u t + 12 g t² solve the quadratic, keep t > 0
Free fall
g = 9.8 m s−2
Exam value
g ≈ 10 m s−2
Fall from rest
v = √(2gh)
1 s of fall
4.9 m
Successive seconds
1 : 3 : 5 …
Reaction time
≈ 0.2 s
V

Relative motion & variable acceleration

ONE FRAME AT A TIME
I.Relative velocity  vAB = vA − vB
II.Relative acceleration  aAB = aA − aB zero for two projectiles
III.Closing on a gap d  t = dvAB
IV.a = a(t)  v = u + ∫ a dt
V.a = a(x)  v dv = a dx
VI.a = a(v)  t = ∫ dva(v) the resisted-motion route
Where marks are lost in this chapter
1“40 km/h out, 60 back, average 50.” No — total distance ÷ total time gives 48, and the average velocity is zero.
2Adding components arithmetically. Speed is √(vx² + vy²), never vx + vy.
3Calling the velocity zero at the top of a throw. Only vy is zero; vx survives untouched.
4Writing all three equations and hoping. Name the variable you neither know nor need, and use the one that omits it.