Part of ME-02 — Kinematics

Exam Cheat Sheet — Kinematics

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Equations of Motion

EquationOmits
v=u+atv = u + ats
s=ut+12at2s = ut + \frac{1}{2}at^2v
v2=u2+2asv^2 = u^2 + 2ast
sn=u+a(2n1)2s_n = u + \frac{a(2n-1)}{2}— (nth second)

Projectile Motion

T=2usinθgH=u2sin2θ2gR=u2sin2θgT = \frac{2u\sin\theta}{g} \qquad H = \frac{u^2\sin^2\theta}{2g} \qquad R = \frac{u^2\sin 2\theta}{g}

  • Max range at θ = 45°: Rmax=u2gR_{max} = \frac{u^2}{g}
  • Complementary angles → same R, H2H1=tan2θ\frac{H_2}{H_1} = \tan^2\theta
  • At peak: v = u cosθ (not zero); vy=0v_y = 0

Circular Motion

ω=vr[rad/s]ac=v2r=ω2r[m/s2, toward centre]\omega = \frac{v}{r} \quad [\text{rad/s}] \qquad a_c = \frac{v^2}{r} = \omega^2 r \quad [\text{m/s}^2, \text{ toward centre}]

Graph Rules

  • x-t slope = velocity | v-t slope = acceleration | v-t area = displacement

Critical Values

  • g=9.810g = 9.8 \approx 10 m/s2s^{2} | sin30°=0.5\sin 30° = 0.5, cos30°=0.866\cos 30° = 0.866, sin45°=0.707\sin 45° = 0.707, sin60°=0.866\sin 60° = 0.866

Sign Convention

  • Fix one direction as positive; maintain throughout entire problem
  • Free fall, upward positive: g = −9.8 m/s2s^{2}

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