AshPhys

Uniform and non-uniform acceleration

AshPhys · Revision cheat sheet
AS Level9702 AS Ch.2Chapter 2: Accelerated motionLesson 2.10Sections 2.10

Goal: Tell uniform from non-uniform acceleration on a graph, and handle a curved v–t graph.

01The big picture

Uniform acceleration means a constant acceleration: a straight v–t graph. Only then can the suvat equations be used.

Non-uniform acceleration gives a curved v–t graph. The acceleration at an instant is the gradient of the tangent there; the displacement is still the area under the curve, estimated by counting squares.

Uniform acceleration [paraphrase]
Constant acceleration: equal changes in velocity in equal times.
Non-uniform acceleration [paraphrase]
The acceleration changes with time; the v–t graph curves.
Fig. 1 · Non-uniform acceleration
Non-uniform accelerationVelocity against time. A curve rises steeply from the origin and levels off towards 12 m/s. A dashed tangent touches it at 3 s. A thin dashed straight line from the origin shows uniform acceleration for comparison.v / m s⁻¹t / suniformnon-uniform31212
The curve flattens: the acceleration falls. The dashed tangent’s gradient is the acceleration at 3.0 s.

02Key equations

Acceleration at an instant
a=gradient of the tangenta = \text{gradient of the tangent}
Displacement
s=area under the curves = \text{area under the curve}
Symbols and units
  • aaaccelerationm s−2\mathrm{m\,s^{-2}}
  • ssdisplacementm\mathrm{m}

03Sample problems with solutions

Sample 1 · Tangent to a curve
In Fig. 1 the tangent at 3.0 s passes through (1.0 s, 4.6 m s⁻¹) and (5.0 s, 10.5 m s⁻¹). Find the acceleration at 3.0 s, and say why v=u+atv = u + at cannot be used here.

  1. Gradient of the tangent:

    a=10.5−4.65.0−1.0=1.48 m s−2a = \frac{10.5 - 4.6}{5.0 - 1.0} = 1.48\ \mathrm{m\,s^{-2}}

  2. The graph curves, so the acceleration is not constant: suvat does not apply.

Answer: a=1.5 m s−2a = 1.5\ \mathrm{m\,s^{-2}} at 3.0 s; acceleration not constant, so no suvat

04Practice problems

  1. P1
    A v–t curve gets less steep as time goes on. Describe the acceleration.
    Final answerPositive but decreasing
    Answer: Positive but decreasing
  2. P2
    A tangent passes through (2.0 s, 3.0 m s⁻¹) and (6.0 s, 15 m s⁻¹). Find the acceleration.
    Final answer3.0 m s⁻²
    Answer: 3.0 m s⁻²
  3. P3
    Under a curved v–t graph there are 46 squares, each 1.0 m s⁻¹ by 0.50 s. Estimate the displacement.
    Final answer23 m
    Answer: 23 m
  4. P4
    Can v2=u2+2asv^2 = u^2 + 2as be used for a skydiver before terminal velocity?
    Final answerNo: the acceleration is not constant
    Answer: No: the acceleration is not constant

05Other useful information

Common mistakes

  • ✗ Wrong: Join two points on the curve to find the acceleration at an instant.

    ✓ Right: That is an average; draw the tangent at the instant.

Exam tips

  • Draw the tangent long, then use a large triangle on it.

What the syllabus asks

  • 9702 topic 2: gradient and area of a v–t graph. Coursebook 2.10.

Explore: Practice questions for this lesson