Part of ME-02 — Kinematics

Error Analysis — Common Kinematics Mistakes

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MistakeWhy It's WrongCorrection
Setting velocity at highest point = 0Only vertical component is zero; horizontal component ucosθu\cos\theta persistsAt peak: v=ucosθv = u\cos\theta (minimum speed, not zero)
Using g = +9.8 when upward is positiveSign convention violated; g acts downwardIf upward +ve, g = −9.8 m/s2s^{2}; if downward +ve, g = +9.8 m/s2s^{2}
Confusing distance with displacementPath of 10 m in a semicircle has displacement = diameter, not 10 mDisplacement = straight-line separation between start and end
Applying SUVAT when acceleration is not constantSUVAT derived assuming a = const; invalid for variable accelerationUse integration: v=adtv = \int a\,dt or graphical area method
Range of 30° > range of 60° (same u)Both are complementary; ranges are equalR30=R60=u2sin60°gR_{30} = R_{60} = \frac{u^2\sin 60°}{g}
s_n formula gives total displacementsns_n is displacement DURING the nth second onlyTotal displacement after n seconds = un+12an2un + \frac{1}{2}an^2
Forgetting 12\frac{1}{2} in s=ut+12at2s = ut + \frac{1}{2}at^2Writing s=ut+at2s = ut + at^2 leads to factor-of-2 errorAlways write 12at2\frac{1}{2}at^2; derive from s=s = area under v-t triangle
Equating angular velocity (rad/s) to linear velocity (m/s)Different dimensions: [ω]=T1[\omega] = T^{-1} vs [v]=LT1[v] = LT^{-1}Use v=ωrv = \omega r to convert between them
Area under x-t graph = displacementArea under x-t graph has dimension [L][T][L][T] — not displacementArea under v-t graph = displacement; slope of x-t graph = velocity
Using sin(2θ) = 2sinθIncorrect trig identitysin2θ=2sinθcosθ\sin 2\theta = 2\sin\theta\cos\theta; for θ = 45°: sin 90° = 1

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