01 · Check the roots
Verify both roots by substitution.
Hint
Use 12t − t².
Worked solution
At t = 2, s = 24 − 4 = 20 m. At t = 10, s = 120 − 100 = 20 m.
Understand · explore · practise
Quadratic times and repeated positions
Solve quadratic time equations, interpret repeated positions and reject roots outside the stated motion interval.
Before you startSUVAT, quadratic equations and signed velocity.
01 / Retain every root until you interpret it
Two positive times can both be physically meaningful.
A particle can pass the same position on its outward and return journeys.
Use s = ut + ½at² to obtain a quadratic in t. Solve it completely, then check the time interval, direction and any event that ends the model. Do not discard the larger root just because it comes second.
One position, different timesExplore
A particle has displacement s = 12t − t² metres for 0 ≤ t ≤ 13 seconds. Choose a target displacement and compare its intersections with the curve.
02 / Solve for an original target
This model has u = 12 m/s and a = −2 m/s².
Find when s = 20 m.Worked example
20 = 12t − t²
Substitute the target displacement.
t² − 12t + 20 = 0
Put the quadratic in standard form.
(t − 2)(t − 10) = 0
Factorise.
t = 2 s or t = 10 s
Both lie in 0 to 13 s; retain both.
02 · Interval between visits
How long passes between the two visits to 20 m?
Hint
Subtract the earlier time from the later.
Worked solution
10 − 2 = 8 s. The interval between visits is different from either clock reading.
03 / Use velocity to distinguish the visits
The displacement value alone does not reveal direction.
Watch: the same position on outward and return motion
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 · Velocity at each visit
Find velocity at the two visits to 20 m.
Hint
v = u + at = 12 − 2t.
Worked solution
At t = 2, v = +8 m/s. At t = 10, v = −8 m/s. The particle passes rightward, then leftward.
04 · Turnaround
When and where does the particle turn?
Hint
Set v = 0, then find s.
Worked solution
12 − 2t = 0 gives t = 6 s. Its displacement is 72 − 36 = 36 m.
04 / A maximum can produce one repeated root or no roots
Connect the discriminant to intersections.
05 · Maximum target
Solve for the time at s = 36 m.
Hint
Complete the square or factorise.
Worked solution
t² − 12t + 36 = (t − 6)² = 0. There is one distinct time, t = 6 s, corresponding to a repeated root.
06 · Unreachable target
Can the particle reach s = 40 m under this model?
Hint
Its maximum displacement is 36 m.
Worked solution
No. The quadratic t² − 12t + 40 = 0 has discriminant 144 − 160 = −16, so there are no real times.
05 / Zero and negative roots need context
A zero time may be an initial event, not a later return.
07 · Back to the start
Solve for s = 0 and identify the later return time.
Hint
Factorise t(12 − t).
Worked solution
t = 0 is the start; t = 12 s is the later return. The time taken to return after departure is 12 s.
08 · Negative target
When is s = −13 m for 0 ≤ t ≤ 13?
Hint
Solve t² − 12t − 13 = 0.
Worked solution
(t − 13)(t + 1) = 0 gives t = 13 or t = −1. Only 13 s lies in the stated interval.
09 · Negative root meaning
Must a negative root mean the algebra was wrong?
Hint
The equation may extend to times before the chosen start.
Worked solution
No. It may describe an extrapolated earlier event, but it is outside a model stated only for t ≥ 0 and must be rejected for that question.
06 / A positive root can still be invalid
Check the model end as well as t ≥ 0.
10 · Changed model interval
Suppose the same motion law is valid only for 0 ≤ t ≤ 8 s. Which visits to 20 m may be reported from it?
Hint
Check t = 2 and t = 10 against the upper bound.
Worked solution
Only t = 2 s is supported. The t = 10 root lies outside the stated interval; later motion needs a new model.
11 · Braking vehicle
A vehicle starts at 12 m/s and brakes at −2 m/s² until rest, then stays still. Is its later visit to 20 m at t = 10 valid?
Hint
Braking ends at t = 6 s and position 36 m.
Worked solution
No. The quadratic continuation would reverse direction. Under the stated stopping model the vehicle stays at 36 m, so only the outward visit at t = 2 s occurs.
07 / Keep exact roots until the final interpretation
Round times only after calculating any further quantities.
12 · Non-integer roots
Find both times at s = 30 m.
Hint
Use (t − 6)² = 6.
Worked solution
t = 6 ± √6 s, approximately 3.55 s and 8.45 s. Both lie in the model interval.
13 · Velocities without early rounding
Find velocities at the exact roots in question 12.
Hint
Substitute 6 ± √6 into 12 − 2t.
Worked solution
The earlier velocity is +2√6 m/s and the later velocity is −2√6 m/s, approximately ±4.90 m/s.
14 · Time spent beyond a position
How long is s > 30 m during this model?
Hint
The parabola lies above the target between its two intersections.
Worked solution
Duration = (6 + √6) − (6 − √6) = 2√6 s, approximately 4.90 s. The units are seconds even though the same numerical value occurs in question 13.
08 / Solve the algebra, then apply the model
Report the event the question actually requests.
Keep both roots where appropriate. Check the permitted time interval and any impact, stop or change of acceleration. Use velocity to label direction at each visit; distinguish the first arrival, later return and time between crossings.
Section 1 of 8 · Retain every root until you interpret it