Which equation of motion should I use?

There's a handy trick for choosing which of the equations of motion, or SUVAT equations, you need to use given a set of terms – let’s call it the VUAST trick. For these kinds of questions, there are always 3 parameters available from which a 4^th can be calculated. Usually, the parameters and their values (or letter, if unknown) are written out as follows:

s = distance [travelled since start]

u = initial velocity

v = final velocity

a = acceleration (must be constant!)

t = time [passed since start]

The equations of motion, which can be derived via a mixture of physical principles, algebra, graphical methods, dimensional analysis and calculus (we’ll save this for a tutorial!), are:

v = u + at {1}

s = ½*(u+v)t {2}

s = ut + ½*at² {3}

v² = u² + 2as {4}

For Maths, these equations must be remembered; for Physics, they appear on the Data Sheet. It turns out that, instead of writing out SUVAT in a column, writing VUAST makes choosing the correct equation much easier. For the top 4 terms, v-u-a-s, use equation {4}; for the bottom 4 terms, u-a-s-t, use equation {3}; for the top 2 and bottom 2 terms, v-u-s-t, use equation {2}. Sadly, the neatness breaks down for equation {1}, but it’s {3} and {4} which cause the most difficulties...

The VUAST trick, while simple, is great for jogging your memory, saving you that little bit of time and averting the panic of a memory blank (we all have them, fear not!).

Bonus: the whole lot – equations, VUAST etc. – work not just for constant linear acceleration, but for constant angular acceleration too, though ΩΩ₀ΑΘT isn’t quite as catchy:

v → ω (final angular velocity)

u → ω₀ (initial angular velocity)

a → α (angular acceleration)

s → θ (angle subtended since start)

t → t (time)

Example (Edexcel M1 Jan 12 Q5):

A stone is projected vertically upwards from a point A with speed u ms^–1. After projection the stone moves freely under gravity until it returns to A. The time between the instant that the stone is projected and the instant that it returns to A is 3⁴/₇ seconds. Modelling the stone as a particle,

(a) show that u = 17½ [3 marks]

v   –u ms^-1

u   +u ms^-1

a   –9.81 ms^-2 (gravity)

s

t   3⁴/₇=²⁵/₇ s

v= –u since “moves freely” suggests no air resistance, so final and initial speeds at A are the same but in opposite directions.

v = u + at → –u = u – 9.81*²⁵/₇ → u = 17.52 ms^-1 q.e.d.

OR

v   0 ms^-1

u   +u ms^-1

a   –9.81 ms^-2 (gravity)

s

t   (3⁴/₇)/2=²⁵/₁₄ s

v=0 at the top of travel, hence the time is halved too.

v = u + at → 0 = u – 9.81*²⁵/₁₄ → u = 17.52 ms^-1 q.e.d.

(b) find the greatest height above A reached by the stone [2]

v   0 ms^-1

u   17.52 ms^-1

a   –9.81 ms^-2

s   s m

t

Top 4 terms → v² = u² + 2as → 0 = 17.52² + 2*–9.81*s → s = 15.64 m

OR

v

u   17.52 ms^-1

a   –9.81 ms^-2

s   s m

t   (3⁴/₇)/2=²⁵/₁₄ s

Bottom 4 terms → s = ut + ½at² → s = 17.52²⁵/₁₄ + ½–9.81(²⁵/₁₄)² → s = 15.64 m

(c) find the length of time for which the stone is at least 6³/₅ m above A. [6]

v

u   17.52 ms^-1

a   –9.81 ms^-2

s   6³/₅=6.6 m

t   t s

Bottom 4 terms → s = ut + ½at² → 6.6 = (17.52)t + ½–9.81t² → (½–9.81)t² + (17.52)t – 6.6 = 0

Solving using the quadratic formula or the Equation function on my recommended calculator, the Casio fx-991ES PLUS:

t = 0.428, 3.143 s which are the times at which the stone crosses the point at s = 6.6 m from A

Hence the time for which s ≥ 6.6 m is given by 3.143 – 0.428 = 2.72 s (check: 2.72 < 3⁴/₇ s)