Constant Acceleration - 2D (Edexcel A-Level Mathematics): Flashcards

There is no such thing as 'squared' vectors

$\mathbf{v} = \mathbf{u} + \mathbf{a}t$

$\mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2$

$\mathbf{s} = \frac{1}{2}(\mathbf{u}+\mathbf{v})t$

$\mathbf{s} = \mathbf{v}t - \frac{1}{2}\mathbf{a}t^2$

$v^2 = u^2 + 2as$ (contains squared vectors)

$y = 0$ and $x$ has a value

$x$-coordinate equals $y$-coordinate

$|\mathbf{v}| = \sqrt{x^2 + y^2}$

Position = Initial position + Displacement

Position vectors equal at the same time

$\mathbf{a} = (0, 0)$ or zero vector