Kinematics in a straight line

Model motion with constant acceleration using the suvat equations and interpret the signs of velocity and acceleration.

About 3 min read
  • Define displacement, velocity and acceleration
  • Choose the correct constant acceleration equation
  • Solve problems that include motion under gravity
Before you start

You can rearrange formulae and solve quadratic equations. You also know that speed is distance divided by time.

Describing motion

Kinematics describes how objects move without asking what causes the motion. In a straight line we choose one direction as positive. Displacement and velocity then carry a sign, so a negative velocity means the object is moving in the opposite direction.

Definition: displacement

The distance from a starting point in a given direction.

Definition: acceleration

The rate of change of velocity, measured in metres per second squared.

The suvat equations

When acceleration is constant, five quantities are linked: displacement s, initial velocity u, final velocity v, acceleration a and time t.

Maths
v = u + at
s = ut + ½at²
s = ½(u + v)t
v² = u² + 2as

List the values you know, identify the one you need, and choose the equation that leaves out the unwanted quantity.

An accelerating car

A car travelling at 5 m/s accelerates at 2 m/s² for 6 s. Find its final velocity and the distance travelled.

Maths
u = 5, a = 2, t = 6
v = u + at = 5 + 2 × 6 = 17 m/s
s = ut + ½at² = 5 × 6 + ½ × 2 × 6² = 66 m

Answer: 17 m/s and 66 m.

Check the conditions

The suvat equations work only when acceleration is constant. If an object moves under gravity, take a = 9.8 m/s² downwards and keep the signs consistent.

Exam focus
Say it like this“Taking up as positive: u = 14, a = −9.8, v = 0”
Not like this“u = 14, a = 9.8, v = 0” — State the positive direction, because gravity acts against upward motion.
Common mistakeMixing signs by treating upward velocity and downward acceleration as both positive.

Practice

A train slows from 30 m/s to 10 m/s in 40 s. Find its acceleration.
a = (v − u) ÷ t = (10 − 30) ÷ 40 = −0.5 m/s².
A cyclist accelerates from rest at 1.5 m/s² for 8 s. Find the distance travelled.
s = ut + ½at² = 0 + ½ × 1.5 × 8² = 48 m.
A car moving at 20 m/s stops with deceleration 4 m/s². Find the stopping distance.
v² = u² + 2as gives 0 = 400 − 8s, so s = 50 m.
Going further
A ball is thrown upwards at 14 m/s. Find the time to reach its highest point and the maximum height.
At the top v = 0. Time: 0 = 14 − 9.8t, so t ≈ 1.43 s. Height: 0 = 196 − 19.6s, so s = 10 m.

Quick check

Kinematics quick check

Which equation does not include time?

It links speeds, acceleration and displacement, so use it when t is neither given nor needed.

An object starts from rest. Which value is known immediately?

From rest means its initial velocity has no size at the start.

What does a negative velocity show?

The sign of velocity gives direction, while a separate acceleration shows how velocity changes.
Key takeaways
  • Choose a positive direction before you start.
  • Use the suvat equations only when acceleration is constant.
  • Pick the equation that leaves out the unwanted quantity.
  • Under gravity, keep the signs consistent.
Back to Grade 12 chaptersRelated resources →