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CIE A-Level Maths Study Notes

3.2.4 Equations of Motion for Constant Acceleration

This section explores the fundamental formulae known as SUVAT equations, which are utilized for solving problems involving uniformly accelerated motion in a straight line. These equations are crucial for understanding the kinematics of various objects, ranging from cars in motion to objects in free fall, under constant acceleration.

Understanding SUVAT Equations

SUVAT equations relate 5 key elements: displacement ss, initial velocity uu, final velocity vv, acceleration aa, and time tt. They're used when acceleration is constant.

The Five Equations:

1. Final Velocity: v=u+atv = u + at

  • Finds final velocity from initial velocity, acceleration, and time.

2. Displacement 1: s=ut+12at2s = ut + \frac{1}{2}at^2

  • Calculates displacement using initial velocity, acceleration, and time.

3. Displacement 2: s=u+v2ts = \frac{u+v}{2} \cdot t

  • Another way to find displacement, using average velocity and time.

4. Velocity-Squared: v2=u2+2asv^2 = u^2 + 2as

  • Connects velocity squares with displacement and acceleration.

5. Displacement 3: s=vt12at2s = vt - \frac{1}{2}at^2

  • A displacement formula using final velocity and time.

Application in Problem-Solving

Example 1: Finding Final Velocity

Question: A car goes from rest to 3m/s23 \, \text{m/s}^2 acceleration for 44 seconds. What's its final velocity?

Solution:

1. Known Values:

  • Initial velocity (u):0m/s(u): 0 \, \text{m/s} (rest).
  • Acceleration (a):3m/s2(a): 3 \, \text{m/s}^2.
  • Time ((t)): 44 seconds.

2. Formula:

  • v=u+atv = u + at.

3. Substitute:

  • v=0+3×4v = 0 + 3 \times 4.

4. Calculate:

  • v=12m/sv = 12 \, \text{m/s}.

5. Answer:

  • Final velocity: 12m/s12 \, \text{m/s}.
Final Velocity vs Time

Example 2: Calculating Displacement

Question: How far does the car move in those 4 seconds?

Solution:

1. Known Values:

  • Initial velocity (u):0m/s(u): 0 \, \text{m/s} (rest).
  • Acceleration (a):3m/s2(a): 3 \, \text{m/s}^2.
  • Time (t)(t): 44 seconds.

2. Formula:

  • s=ut+12at2s = ut + \frac{1}{2}at^2.

3. Substitute:

  • s=0×4+12×3×42s = 0 \times 4 + \frac{1}{2} \times 3 \times 4^2.

4. Calculate:

  • s=0+24=24ms = 0 + 24 = 24 \, \text{m}.

5. Answer:

  • Displacement: 24m24 \, \text{m}.
Displacement vs Time
Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
LinkedIn
Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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