Content
Assumptions for SUVAT
The equations apply to straight-line motion with constant acceleration. Air resistance and changing forces are neglected unless stated.
SUVAT equations only work when acceleration is constant. If acceleration varies, you must use calculus or graphical methods.
Core equations
The four SUVAT equations are:
\[v = u + at\]
\[s = ut + \frac{1}{2}at^2\]
\[v^2 = u^2 + 2as\]
\[s = \frac{(u+v)}{2}t\]
Where:
- \(s\) = displacement (m)
- \(u\) = initial velocity (m/s)
- \(v\) = final velocity (m/s)
- \(a\) = acceleration (m/s²)
- \(t\) = time (s)
Equation Selection Guide
| \(s\) |
\(v = u + at\) |
| \(v\) |
\(s = ut + \frac{1}{2}at^2\) |
| \(u\) |
\(s = vt - \frac{1}{2}at^2\) |
| \(a\) |
\(s = \frac{(u+v)}{2}t\) |
| \(t\) |
\(v^2 = u^2 + 2as\) |
Problem-solving strategy
- List known variables with units.
- Identify the unknown variable.
- Choose an equation that includes the knowns and the unknown.
- Substitute, solve, and check sign conventions.
Interactive: SUVAT Equation Solver
Use this interactive tool to solve kinematics problems. Enter any three known values, and the solver will calculate the remaining two.
- Enter at least 3 known values (leave unknown fields empty)
- Click “Solve” to calculate the missing values
- The tool shows which equation was used
Worked Examples
Example 1: Final velocity
A cyclist starts at 3.0 m/s and accelerates at 0.80 m/s\(^2\) for 5.0 s.
Known: \(u = 3.0\) m/s, \(a = 0.80\) m/s², \(t = 5.0\) s Unknown: \(v\)
- Use \(v = u + at\)
- \(v = 3.0 + (0.80)(5.0) = 3.0 + 4.0\)
- \(v = 7.0\) m/s
Velocity increases in the positive direction.
Example 2: Displacement under acceleration
A car travels at 10 m/s and accelerates at 1.5 m/s\(^2\) for 4.0 s.
Known: \(u = 10\) m/s, \(a = 1.5\) m/s², \(t = 4.0\) s Unknown: \(s\)
- Use \(s = ut + \frac{1}{2}at^2\)
- \(s = (10)(4.0) + \frac{1}{2}(1.5)(4.0)^2\)
- \(s = 40 + 12 = 52\) m
Example 3: Stopping distance
A train slows uniformly from 22 m/s to rest over 110 m. Find the acceleration.
Known: \(u = 22\) m/s, \(v = 0\) m/s, \(s = 110\) m Unknown: \(a\)
- Use \(v^2 = u^2 + 2as\) (no \(t\) needed)
- \(0^2 = 22^2 + 2a(110)\)
- \(0 = 484 + 220a\)
- \(a = -2.2\) m/s²
The negative sign indicates deceleration.
Example 4: Free fall
A ball is dropped from rest. How far does it fall in 3.0 s? (Take \(g = 9.8\) m/s² downward)
Known: \(u = 0\) m/s, \(a = 9.8\) m/s², \(t = 3.0\) s Unknown: \(s\)
- Use \(s = ut + \frac{1}{2}at^2\)
- \(s = 0 + \frac{1}{2}(9.8)(3.0)^2\)
- \(s = 44\) m
Visualization: Motion Under Constant Acceleration
The graphs below show the relationship between position, velocity, and time for an accelerating object:
Observe:
- The curve is a parabola (characteristic of constant acceleration)
- The gradient increases over time (velocity increases)
- The tangent line at any point gives instantaneous velocity