Kinematics & Arithmetic Series – 9 July Class Notes

Arithmetic Progressions (A.P.)

• Key quantities
• First term aa
• Common difference dd
• n-th term a<em>na<em>n • Sum of first nn terms S</em>nS</em>n

• Formulae (as repeatedly written in the pages)
• n-th term: a<em>n=a+(n1)da<em>n = a + (n-1)d • Sum of first n terms – two equivalent layouts were scribbled • Standard: S</em>n=n2[2a+(n1)d]S</em>n = \dfrac{n}{2}\,[2a+(n-1)d]
• Alternative (seen in transcript): S<em>n=a+12d(2n1)nS<em>n = a + \dfrac{1}{2}d\,(2n-1)\,n — same expression written more compactly as S</em>n=4+1a(2n1)S</em>n = 4 + 1a(2n-1) for the worked example where a=4a = 4.

• Worked sequence snippets found on Page 1
• First terms recorded: 4,  ?,19,24,\;? ,19,2 – the illegible “+ +” indicates at least four terms were being listed.
• Determining an n-th term
• Equation copied: 24=4(4)+ja(4)224 = 4(4) + ja(4)2 → interpreted as a<em>n=24=4+(n1)da<em>n = 24 = 4 + (n-1)d where n=4n=4 yields d=203d=\dfrac{20}{3}. • Sum for n=8n=8 noted twice • Line written: S</em>8=4(+)+19+2S</em>8 = 4(+) + 19 + ^2 → interpreted as S<em>8=4+(1/2)d(281)8S<em>8 = 4 + (1/2)d(2·8-1)·8. • Numerical result seen: 88=4(8)+59(8)288 = 4(8) + 59(8) 2 → points to S</em>8=88S</em>8 = 88.
• Additional numeric endpoints scattered on Page 1: 23,24,26,29,49,324{23, 24, 26, 29, 49, 324} – these are residual sums/terms from trial values of nn.

Straight-Line Kinematics (SUVAT)

Fundamental relations

• Velocity–time: v=u+atv = u + at
• Displacement–time: s=ut+12at2s = ut + \dfrac{1}{2}at^{2}
• Velocity–displacement: v2=u2+2asv^{2} = u^{2} + 2as
• Average acceleration: a=Δvt=v<em>2v</em>1ta = \dfrac{\Delta v}{t} = \dfrac{v<em>2-v</em>1}{t}

Page 1 worked example (sign-change case)

• Given: initial velocity v<em>1=10m/sv<em>1 = 10\,\text{m/s}, final velocity v</em>2=2m/sv</em>2 = -2\,\text{m/s}, time t=4st = 4\,\text{s}.
• Change in velocity: Δv=v<em>2v</em>1=210=12m/s\Delta v = v<em>2 - v</em>1 = -2 - 10 = -12\,\text{m/s}.
• Acceleration: a=124=3m/s2a = \dfrac{-12}{4} = -3\,\text{m/s}^2 (recorded as “az = –3 m/ s²”, though the draft shows multiple crossed-out values of –6 and –8×10⁻⁴ before settling).

Page 2 two-axis projectile style problem

• Scenario split into X and Y components for the first 4 s of motion.

X-axis data

• Initial velocity: u<em>x=0u<em>x = 0. • Acceleration: a</em>x=6m/s2a</em>x = 6\,\text{m/s}^2 (written “An = 6 m/see²”).
• Time: t=4st = 4\,\text{s}.
• Displacement: s<em>x=u</em>xt+12axt2=12(6)(4)2=48ms<em>x = u</em>xt + \tfrac12 a_xt^2 = \tfrac12(6)(4)^2 = 48\,\text{m}.

Y-axis data (two slightly different versions appear)
  1. Version with zero initial speed (matches numbers on the sheet)
    u<em>y=0u<em>y = 0, a</em>y=8m/s2a</em>y = 8\,\text{m/s}^2.
    sy=12(8)(4)2=64ms_y = \tfrac12(8)(4)^2 = 64\,\text{m}.

  2. Version including an initial speed line “Uy = 20” which, if used, would give
    sy=(20)(4)+12(8)(4)2=80+64=144ms_y = (20)(4) + \tfrac12(8)(4)^2 = 80 + 64 = 144\,\text{m}.
    • Transcript ultimately preserves the 64m64\,\text{m} figure.

Resultant displacement & direction

• Magnitude
R=s<em>x2+s</em>y2=482+642=2304+4096=6400=80mR = \sqrt{s<em>x^{2}+s</em>y^{2}} = \sqrt{48^{2}+64^{2}} = \sqrt{2304+4096} = \sqrt{6400} = 80\,\text{m}.
• Direction above the X-axis
tanθ=s<em>ys</em>x=6448=43θ53\tan\theta = \dfrac{s<em>y}{s</em>x} = \dfrac{64}{48} = \dfrac{4}{3} \Rightarrow \theta \approx 53^{\circ}.
• Angles written: “48°” (round-off) and “α = 53°” appear; 53° is the correct arc-tan value.

Miscellaneous single-axis suvat snapshots

  1. Line: V=9km/suV = 9\,\text{km/su}, u=1km/su = 1\,\text{km/s}, s=4ms = 4\,\text{m}.
    • If treated with v2=u2+2asv^{2}=u^{2}+2asa=v2u22s=8118=10m/s2a = \dfrac{v^{2}-u^{2}}{2s} = \dfrac{81-1}{8} = 10\,\text{m/s}^2.

  2. Equation copied: (10)22a(20)=0(10)^{2} - 2a(20) = 0
    • Rearranged gives a=10040=2.5m/s2a = \dfrac{100}{40} = 2.5\,\text{m/s}^2 (sheet shows “a = 5/2 ≈ 2.5”).

  3. Equation copied: (30)22aS=0(30)^{2} - 2a S = 0 with final S=180mS = 180\,\text{m}.
    • Implies 900=2a(180)900 = 2a(180)a=2.5m/s2a = 2.5\,\text{m/s}^2 again – consistent with the previous line.

Pythagorean & Vector Identities Seen

• Repeated use of 2=42+208\sqrt{2} = 4^{2} + 208 and 42203\sqrt{42 - 203} lines are mis-writes; intended relations are of the form v2=u2+2asv^{2} = u^{2} + 2as or R=X2+Y2R = \sqrt{X^{2}+Y^{2}}.

Numerical Table of Highlighted Results

• Displacements: 48m48\,\text{m} (X), 64m64\,\text{m} (Y), 80m80\,\text{m} (resultant).
• Accelerations explicitly boxed: a=3m/s2a = -3\,\text{m/s}^2, a=2.5m/s2a = 2.5\,\text{m/s}^2, a=10m/s2a = 10\,\text{m/s}^2.
• Angle: θ53\theta \approx 53^{\circ} above the positive X-axis.
• Sum of first 8 terms of the displayed A.P.: S8=88S_8 = 88.

Connections & Context

• Arithmetic-progression manipulation is immediately followed by SUVAT work, illustrating the common examination pattern: Section A (pure maths) then Section B (mechanics).
• Vector displacement problem connects to projectile motion, reinforcing earlier lessons on decomposing motion into orthogonal components and re-combining with Pythagoras and trigonometry.
• Sign conventions (positive/negative velocity) underpin the change-in-velocity example, echoing the emphasis on coordinate direction set in earlier July classes.

Practical / Exam Tips (drawn from annotations)

• Always write units (the sheets oscillate between km/s, m/s, s, m) – consistency avoids a 1-mark penalty.
• Box final answers – many of the numbers (e.g., 80, 64, 48) are underlines or boxed in the transcript.
• Double-check calculator inputs; sheet shows mis-keys like “8 × 10⁻⁴” that were later crossed out.
• For A.P. sums, keep the n/2n/2 factor visible; dropping it led to the erroneous 24 ⇒ 6 conversion on Page 1.
• Use a sign diagram when switching direction (e.g.
v2=2m/sv_2 = -2\,\text{m/s} opposite to u=+10m/su = +10\,\text{m/s}).