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Kinematics in a straight line

Displacement, velocity and acceleration, with the constant-acceleration SUVAT equations.

  • Define and explain Kinematics in a straight line in your own words
  • Use key terms such as Displacement accurately
  • Apply what you have learned to new examples and questions
  • Avoid the common mistakes learners make with this topic

Mechanics is mathematics with muscles: it predicts how real objects move. The same equations that time a sprinter's finish also guide satellites — learn them once and the universe starts making sense.

This lesson focuses on Kinematics in a straight line: displacement, velocity and acceleration, with the constant-acceleration SUVAT equations.

Definition: Kinematics in a straight line

Displacement, velocity and acceleration, with the constant-acceleration SUVAT equations.

Key ideas

SUVAT equations

For constant acceleration: v = u + at, s = ut + ½at², and v² = u² + 2as. Choose the equation containing the quantities you know and the one unknown you need; always define a positive direction first.

Projectile motion

Horizontal and vertical motion are independent: horizontal velocity stays constant (no air resistance), while vertical motion accelerates at g = 9.8 m/s² downwards. Resolve the initial velocity into components, then treat each direction with SUVAT.

Key term — Displacement: Distance measured in a straight line from a starting point, with direction included; unlike distance, it can be negative.

Constant acceleration

A car accelerates uniformly from 4 m/s to 12 m/s in 4 s. Find the acceleration and the distance travelled.

Acceleration a = (v − u)/t = (12 − 4)/4 = 8/4 = 2 m/s². Distance via s = (u + v)/2 × t = (4 + 12)/2 × 4 = 8 × 4 = 32 m. Check with s = ut + ½at² = 16 + 16 = 32 m. ✓.

Answer: Acceleration 2 m/s²; distance 32 m.

Common mistakes
  • Confusing speed with velocity, or distance with displacement SUVAT uses vectors: fix a positive direction at the start and keep signs consistent, or the equations give nonsense.
  • Using F = ma with the wrong force It is the resultant (net) force that equals ma — add all forces vectorially first, including friction and weight components.

Practice

A particle starts from rest and accelerates at 3 m/s² for 5 s. Find its final velocity.
v = u + at.

v = 0 + 3 × 5 = 15 m/s.

A 2 kg mass experiences a resultant force of 10 N. Find its acceleration.
F = ma.

a = 10/2 = 5 m/s².

Find the distance travelled in the previous question.
s = ut + ½at².

s = 0 + ½ × 3 × 25 = 37.5 m.

A box on rough ground has normal reaction 50 N and μ = 0.4. Find the maximum friction.
F_max = μR.

F_max = 0.4 × 50 = 20 N.

Quick check

Kinematics in a straight line — quick check

Which of these best defines "Displacement"?

Distance measured in a straight line from a starting point, with direction included; unlike distance, it can be negative.

A projectile is launched horizontally at 20 m/s from a 45 m cliff. How long until it hits the sea? (g = 9.8 m/s²)

t² = 90/9.8 ≈ 9.18; t ≈ 3.03 s.

A ball is dropped from rest. How far does it fall in 2 s? (g = 9.8 m/s²)

s = ½ × 9.8 × 4 = 19.6 m.
Key takeaways
  • Kinematics in a straight line: displacement, velocity and acceleration, with the constant-acceleration SUVAT equations.
  • SUVAT equations: For constant acceleration: v = u + at, s = ut + ½at², and v² = u² + 2as.
  • Acceleration: The rate of change of velocity, in m/s²; positive for speeding up in the chosen direction, negative (deceleration) for slowing down.
  • Watch out for: confusing speed with velocity, or distance with displacement
Newton's laws of motion →Back to Grade 12 chaptersRelated resources →