Non-formulae:

$a = a_{m} \pm Δ a$ $Δ a = \sum | a_{m} - a_{i} |$
$n VSD = (n - 1) MSD$
$LC = VC = 1 MSD - 1 VSD = \frac{1}{n} MSD$
$LC = \frac{pitch}{number of divisions}$
$| A + B | = \sqrt{{| A |}^{2} + {| B |}^{2} + 2 | A | | B | \cos θ}$
$Angle of A + B with A = \tan^{- 1} (\frac{B \sin θ}{A + B \cos θ})$
$A \cdot B = | A | | B | \cos θ = A_{x} B_{x} + A_{y} B_{y} + A_{z} B_{z}$
$A ⨯ B = | \begin{matrix} i & j & k \\ A_{x} & A_{y} & A_{z} \\ B_{x} & B_{y} & B_{z} \end{matrix} | = | A | | B | \sin θ * direction$
$v = u + a t$
$s = u t + \frac{a t^{2}}{2}$
$v^{2} - u^{2} = 2 a s$
$s_{n} = u + \frac{a (2 n - 1)}{2}$
$H = \frac{{(u \sin θ)}^{2}}{2 g} = \frac{{u_{y}}^{2}}{2 g}$
$R = \frac{u^{2} \sin^{2} θ}{g} = 4 H \cot θ$
$R_{horizontal projectile} = \frac{u_{x} u_{y}}{g} = u_{x} \frac{\sqrt{H}}{\sqrt{2 g}}$
$Minimum speed required to hit (x, y) from (0, 0) = u_{\min} = \sqrt{g (y + \sqrt{x^{2} + y^{2}})}$
$Minimum angle = \tan^{- 1} [\frac{u_{\min}^{2}}{g x}]$

Formulae:

$Zero error = - zero correction$
$Length = reading + zero correction$
${(1 + x)}^{n} = 1 + n x + \frac{n (n - 1) x}{2!} + \frac{n (n - 1) (n - 2) x}{3!} + ..., | x | < 1$
$\frac{a}{\sin (A)} = \frac{b}{\sin (B)} = \frac{c}{\sin (C)}$
$A = | A | \times dir (A)$
$| A | = \sqrt{A_{x}^{2} + A_{y}^{2} + A_{z}^{2}}$
$D_{A|B} = D_{A} - D_{B}$
$t_{f} = \frac{2 (u \sin θ)}{g}$
$t_{d} = \frac{\sqrt{H}}{\sqrt{2 g}}$
$For man in rain to remain dry, he should hold his umbrella at an angle of θ = \tan^{- 1} (\frac{V_{m}}{V_{r}})$
$a = \frac{F}{m}$
$J = \int \sum F ⅆ t = F_{avg} \times t$
$W = m g, g = 9.8 m s^{- 2}$ W is weight or, the pull of the Earth on you and g is the accelaration due to Earth's gravity

Use dot product to find angle between vectors
Verify if graph given is trajectory or distance-time graph
$For projectile at a height, find t_{f} using s = u t + \frac{a t^{2}}{2}$
For finding shortest distance, consider relative motion
For collisions, the trajectories should intersect and at the point of intersection the times must be the same
Drawing FBD is vital to not making mistakes
$Dividing forces into x, y, z components can be helpful in solving questions$