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Deriving the equations of motion

AshPhys · Revision cheat sheet
AS Level9702 AS Ch.2Chapter 2: Accelerated motionLesson 2.9Sections 2.9

Goal: Derive the four suvat equations from a velocity–time graph.

01The big picture

For uniform acceleration the v–t graph is a straight line from uu to vv: its gradient is aa and the area under it is ss.

Those give the first two equations; algebra gives the other two.

Uniformly accelerated motion [paraphrase]
Constant acceleration in a straight line.

02Deriving the equations

D1 From the v–t graph

Velocity rises uniformly from uu to vv in time tt.

Fig. 1 · The area under a uniform v–t line
The area under a uniform v–t lineVelocity against time. A straight line rises from u at time 0 to v at time t. The area under it is split by a dashed line at height u into a rectangle of area ut and a triangle of height v minus u.vtuvtut½(v − u)tv − u
Rectangle utut plus triangle 12(v−u)t\tfrac12(v - u)t; the gradient is aa.

  1. Gradient = acceleration:

    a=v−ut  ⇒  v=u+ata = \frac{v - u}{t} \;\Rightarrow\; v = u + at

  2. Area = rectangle + triangle:

    s=ut+12(v−u)t=12(u+v) ts = ut + \tfrac{1}{2}(v - u)t = \tfrac{1}{2}(u + v)\,t

  3. Put v−u=atv - u = at into the area:

    s=ut+12at2s = ut + \tfrac{1}{2}at^{2}

  4. Put t=(v−u)/at = (v - u)/a into the area:

    s=(v+u)(v−u)2a  ⇒  v2=u2+2ass = \frac{(v + u)(v - u)}{2a} \;\Rightarrow\; v^{2} = u^{2} + 2as
Result
v=u+at,s=12(u+v) t,s=ut+12at2,v2=u2+2as\begin{aligned} v &= u + at, & s &= \tfrac{1}{2}(u + v)\,t, \\ s &= ut + \tfrac{1}{2}at^{2}, & v^{2} &= u^{2} + 2as \end{aligned}

03Sample problems with solutions

Sample 1 · Area check
An object accelerates uniformly from 4.0 m s⁻¹ to 10 m s⁻¹ in 3.0 s. Find ss from the area, then check it with s=ut+12at2s = ut + \tfrac{1}{2}at^2.

  1. Area (trapezium):

    s=12(4.0+10)(3.0)=21 ms = \tfrac{1}{2}(4.0 + 10)(3.0) = 21\ \mathrm{m}

  2. Check: a=(10−4.0)/3.0=2.0a = (10 - 4.0)/3.0 = 2.0 m s⁻², so

    s=4.0(3.0)+12(2.0)(3.0)2=12+9.0=21 m ✓s = 4.0(3.0) + \tfrac{1}{2}(2.0)(3.0)^{2} = 12 + 9.0 = 21\ \mathrm{m}\ \checkmark
Answer: s=21 ms = 21\ \mathrm{m} both ways

04Practice problems

  1. P1
    Which feature of a v–t graph gives v=u+atv = u + at?
    Final answerThe gradient
    Answer: The gradient
  2. P2
    Which suvat equation has no tt, and how is it derived?
    Final answerv2=u2+2asv^2 = u^2 + 2as (eliminate tt)
    Answer: v2=u2+2asv^2 = u^2 + 2as (eliminate tt)
  3. P3
    In s=ut+12at2s = ut + \tfrac12 at^2, which part of the graph area is utut?
    Final answerThe rectangle under uu
    Answer: The rectangle under uu
  4. P4
    Find ss from the area: u=0u = 0, v=12v = 12 m s⁻¹, t=5.0t = 5.0 s.
    Final answer30 m
    Answer: 30 m

05Other useful information

Common mistakes

  • ✗ Wrong: Use suvat for any motion.

    ✓ Right: Only for constant acceleration.

Exam tips

  • Forgot an equation? Sketch the v–t trapezium.

What the syllabus asks

  • 9702 topic 2: derive the suvat equations. Coursebook 2.9.

Explore: Practice questions for this lesson