A quadratic equation gives two roots, and the physical setup may allow only one. Always compare each root with what the object is actually doing, then reject the one that breaks the situation and write why.
This lesson is part of selecting and checking a physical model in SPM Physics. It depends on using equations of uniformly accelerated motion.
Which root is impossible in a braking problem?
A car travelling at 20 m s⁻¹ brakes at a constant 5.0 m s⁻². At what time is it 30 m from where it began to brake?
Take forward as positive: u = 20, a = −5, s = 30. Use s = ut + ½at²: 30 = 20t − 2.5t². Rearranging, 2.5t² − 20t + 30 = 0, so t² − 8t + 12 = 0, which factorises to (t − 2)(t − 6) = 0.
The roots are t = 2 s and t = 6 s. Both satisfy the equation, but only one can describe the car.
Worked example: applying the physical limit
Find the stopping time: v = u + at gives 0 = 20 − 5t, so t = 4 s. After 4 s the car is at rest, and the constant-deceleration model ends.
At t = 6 s the formula would have the car reversing back towards the start, which a braking car does not do. The root t = 6 s is rejected. The valid time is t = 2 s.
A full answer reads: “The car stops at t = 4 s, so t = 6 s is not possible. The car passes 30 m at t = 2 s.” Check: at 2 s, s = 20 × 2 − 2.5 × 4 = 30 m. The car is then still moving at 10 m s⁻¹.
When are both roots valid?
A ball is thrown up at 20 m s⁻¹ from the ground. When is it at a height of 15 m? Take g = 10 m s⁻² with upward positive.
15 = 20t − 5t², so t² − 4t + 3 = 0, which gives (t − 1)(t − 3) = 0. The roots are t = 1 s and t = 3 s.
Both are valid. At 1 s the ball is rising, and at 3 s it is falling. Nothing has stopped the motion before 3 s, because the ball lands at 4 s.
The mistake to avoid
The common mistake is to reject a root because it is larger or because “one answer is expected”. Use a physical reason for each decision.
| Situation | Rejected root | Reason |
|---|---|---|
| Car braking to rest | t = 6 s | The car stopped at 4 s |
| Ball thrown upward | none | Both times occur |
| Any time problem | negative t | The event begins at t = 0 |
The units and significant figure checker cannot reject roots for you, so the physical reasoning has to come from you.
Check yourself
A trolley moving at 6.0 m s⁻¹ slows at 2.0 m s⁻². Solve 8 = 6t − t² for t and say which root is valid.
Answer
Rearrange: t² − 6t + 8 = 0, so (t − 2)(t − 4) = 0. The roots are t = 2 s and t = 4 s.
The trolley stops at t = 3 s, since 6 − 2t = 0. By then it has travelled 6 × 3 − 9 = 9 m. The root t = 4 s would put the trolley back at 8 m, which needs it to reverse, but it stays at rest at 9 m. The valid root is t = 2 s, and t = 4 s is rejected.
What to study next
Test all four checks in the integrated practice set. If you want to revisit boundaries first, read drawing a labelled system boundary before choosing a conservation equation.
If you want a teacher to go through your rejected roots with you, see online one-to-one Physics tuition.