01Why this matters
A formula written for distance can also find time. Rearranging once is clearer than guessing values repeatedly.
02See it
03Understand it
Rearranging a formula makes a different variable the subject while preserving the relationship.
The subject is the variable isolated on one side. Use inverse operations on both sides, respecting brackets and any restrictions on division. A useful check is to substitute simple numbers into the original and rearranged forms and confirm agreement.
04Guided example
Rearrange v = d/t to make t the subject, then find the time for 150 m at 5 m/s.
- Multiply both sides by t: vt = d. This removes t from the denominator.
- Divide both sides by v: t = d/v, provided v is not zero.
- Substitute: t = 150 m ÷ 5 m/s = 30 s.
05Try it
06Practise it
07Use it in the real world
To find the width of a rectangular 24 m² floor with length 6 m, use w = A/l = 24/6 = 4 m.
Go a little deeper
When the unknown occurs in several terms, factor it out first. For example, ax + bx = c becomes x(a + b) = c, then x = c/(a + b) if a + b ≠ 0.
08Quick check
09Summary
Rearranging a formula makes a different variable the subject while preserving the relationship. Substitute: t = 150 m ÷ 5 m/s = 30 s.
Before moving on, explain the worked example in your own words. Use the hint on any question you want to revisit.
10Key idea
v = d/t ↔ t = d/v
v is constant speed in m/s; d is distance in m; t is time in s.
d = vt; t = d/v for v ≠ 0
Find this in the formula library →