01 · Reaction distance
At 20 m/s with reaction time 0.75 s, how far does the vehicle travel before braking?
Hint
Speed stays constant during reaction.
Worked solution
Reaction distance = 20 × 0.75 = 15 m.
Understand · explore · practise
Constant acceleration: mixed practice
Practise reaction and braking distances, multistage journeys, parameter models, reversals and vertical motion with independent worked solutions.
Before you startThe constant-acceleration chapter; attempt each question before revealing its hint.
01 / Choose the model before the equation
Use a sketch, a clock and a sign convention.
Work through each problem independently first.
Identify constant-acceleration stages, launch delays, turns and contacts. Keep signed displacement separate from distance. Check a root against the physical time interval before using it in the next step.
Reaction, then brakingExplore
A vehicle travels at 20 m/s during the chosen reaction time, then brakes with constant acceleration −4 m/s² until rest. Its velocity graph ends at stopping; no reversal is assumed. Compare its stopping distance with an obstacle 60 m ahead.
02 / Combine reaction distance and braking distance
The vehicle moves before braking begins.
Watch: reaction adds a rectangle to the stopping area
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 · Braking stage
It then brakes at 4 m/s² until rest. Find braking time and distance.
Hint
Use v = u + at and v² = u² + 2as, with a = −4.
Worked solution
Braking time = 5 s. Braking distance = 20²/(2 × 4) = 50 m.
03 · Stopping condition
Will it stop before an obstacle 60 m ahead?
Hint
Add reaction and braking distances.
Worked solution
It needs 15 + 50 = 65 m, so it cannot stop before the obstacle in this model. At 60 m it is still moving.
04 · Speed at the obstacle
Find its speed on reaching that obstacle.
Hint
After reaction it has 45 m available for braking.
Worked solution
v² = 20² − 2(4)(45) = 40, so v = √40 ≈ 6.32 m/s. Use the positive root because the vehicle has not reversed.
05 · Maximum reaction time
Find the largest reaction time that lets it stop by the obstacle, with all other data unchanged.
Hint
The braking distance is 50 m.
Worked solution
20r + 50 ≤ 60 gives r ≤ 0.5 s. Stopping strictly before it requires r < 0.5 s.
03 / Locate an event within the correct journey stage
A midpoint in distance need not occur halfway through time.
06 · Three-stage journey
A particle starts from rest, accelerates at 2 m/s² for 6 s, travels at constant speed for 8 s, then decelerates at 3 m/s² until rest. Find its maximum speed, total time and distance.
Hint
Draw a trapezium-shaped velocity graph.
Worked solution
Maximum speed = 12 m/s. Braking takes 4 s; total time = 18 s. Distances are 36 m, 96 m and 24 m, so total distance = 156 m.
07 · Halfway in distance
For question 6, find when it has travelled half the total distance.
Hint
Half is 78 m; identify the stage containing that point.
Worked solution
After 6 s it has travelled 36 m. It needs another 42 m at 12 m/s, taking 3.5 s. The halfway point is at overall t = 9.5 s.
08 · Halfway in time
Find the distance travelled by t = 9 s in that journey and its average speed over the complete journey.
Hint
Nine seconds includes 3 s of the constant-speed stage.
Worked solution
Distance at 9 s = 36 + 12(3) = 72 m. Overall average speed = 156/18 = 26/3 ≈ 8.67 m/s.
04 / Use the total distance to determine an unknown acceleration
Keep the braking stage dependent on the earlier speed.
09 · Write the distance equation
A particle accelerates from rest at k m/s² for 5 s, cruises for 4 s, then brakes at 2 m/s² until rest. Here k > 0. Its total distance is 90 m. Form an equation for k.
Hint
Its maximum speed is 5k; the braking distance is (5k)²/4.
Worked solution
Acceleration distance = 12.5k, cruise distance = 20k, braking distance = 6.25k². Thus 6.25k² + 32.5k = 90, or 5k² + 26k − 72 = 0.
10 · Solve and interpret
Find k and the total journey time for question 9.
Hint
Factor (k − 2)(5k + 36).
Worked solution
The roots are k = 2 and k = −7.2; reject the negative root because k > 0. Maximum speed is 10 m/s, braking lasts 5 s, and total time is 5 + 4 + 5 = 14 s.
05 / A signed area can hide a return journey
Split distance where velocity changes sign.
11 · Turning time
A particle moves with u = 8 m/s and a = −2 m/s² for 6 s. Find when it reverses and its final velocity.
Hint
Solve v = 0, then calculate v at 6 s.
Worked solution
v = 8 − 2t, so it reverses at 4 s. Final velocity is −4 m/s.
12 · Distance and displacement
Find both quantities over those 6 s.
Hint
Split the journey at 4 s.
Worked solution
Displacement = 8(6) − 6² = 12 m. Outward distance is 16 m and return distance is 4 m, so total distance = 20 m.
13 · Average quantities
Find average velocity and average speed over the same 6 s.
Hint
Use displacement or distance as appropriate.
Worked solution
Average velocity = 12/6 = 2 m/s. Average speed = 20/6 = 10/3 ≈ 3.33 m/s.
06 / Combine height equations with the flight domain
Use g = 9.8 m/s² and ignore air resistance in these problems.
14 · Two height crossings
A particle launches upward at 24.5 m/s from ground. When is it 29.4 m high?
Hint
Solve 24.5t − 4.9t² = 29.4.
Worked solution
t² − 5t + 6 = (t − 2)(t − 3) = 0. Both t = 2 s and t = 3 s occur before return to ground at 5 s.
15 · Above the target
For how long is that particle strictly above 29.4 m, and what is its maximum height?
Hint
Use the interval between the roots, then v = 0 at the apex.
Worked solution
It is above for 2 < t < 3, lasting 1 s. Maximum height = 24.5²/19.6 = 30.625 m.
16 · Delayed identical launch
A second particle launches from ground 1 s later with the same upward speed 24.5 m/s. Find their meeting time and height.
Hint
Use t for the first particle and t − 1 for the second.
Worked solution
Equating heights cancels the quadratic terms and gives 9.8t = 29.4, so overall t = 3 s. Meeting height = 29.4 m. The first is descending at −4.9 m/s; the second is ascending at +4.9 m/s.
17 · A rebound
A particle is dropped from 20 m and its upward rebound speed is 0.6 times its pre-impact speed. Find its maximum rebound height.
Hint
Square the speed fraction to obtain the height fraction.
Worked solution
Rebound height = 0.6² × 20 = 7.2 m. This uses the stated impact rule, not a continuation of the drop equation.
07 / Check assumptions and proposed solutions
An algebraic answer must describe the stated event.
18 · Averaging endpoint speeds
May (initial speed + final speed)/2 be used as the average speed for question 6’s whole journey?
Hint
The acceleration changes between stages.
Worked solution
No. Both endpoint speeds are zero, yet the particle travels 156 m. Use total distance divided by total time, or areas for the separate stages.
19 · Braking extrapolation
A stopping model gives a negative velocity when extended past the stopping time. Does this prove the vehicle reverses?
Hint
The stated model only covers braking until rest.
Worked solution
No. Reversal would require an additional assumption. End this braking stage at v = 0 and specify later motion separately.
20 · Gravity and mass
A learner says a heavier particle falls faster because its weight is greater. Explain what the ideal free-fall model predicts.
Hint
Use W = mg and a = W/m, ignoring air resistance.
Worked solution
Weight is greater for greater mass, but dividing by mass gives the same g. Actual air resistance can produce different falls; the ideal prediction depends on the assumptions.
08 / Use your errors to choose what to revisit
Identify whether the difficulty was modelling, algebra or interpretation.
If your equation was wrong, redraw the stages or label the origin and clock. If the equation was right but the answer was wrong, check roots, signs and arithmetic. If the number was right but the conclusion was wrong, distinguish time from duration, distance from displacement, and the valid flight or braking interval.
Section 1 of 8 · Choose the model before the equation