01 · Before the turn
Find velocity, displacement and distance after 2 s.
Hint
Velocity stays positive over this interval.
Worked solution
v = 6 m/s; s = 20 − 4 = 16 m; distance = 16 m.
Understand · explore · practise
Stopping, reversal and total distance
Find a turning time, split the path to calculate total distance and distinguish continued acceleration from braking that ends at rest.
Before you startSigned SUVAT equations and the meaning of distance and displacement.
01 / Find whether velocity changes sign
An instant of zero velocity is not automatically a stationary interval.
Solve v = u + at = 0, then check the interval.
If a nonzero constant acceleration continues through that time, velocity changes sign and the particle reverses. If the model says braking stops at rest, do not extend the same acceleration beyond the stopping time.
Position and distance after a turnExplore
A particle starts at x = 0 with u = 10 m/s and constant a = −2 m/s² throughout. Choose an elapsed time and compare its coordinate with total distance travelled.
02 / Locate the turn before finding distance
The turning point splits the path into one-direction pieces.
For u = 10 m/s, a = −2 m/s², find the turn.Worked example
0 = 10 − 2t
Set velocity to zero.
t = 5 s
This is the potential turning time.
s = 10 × 5 − 5² = 25 m
The particle is 25 m right of its start at the turn.
02 · At the turn
What are velocity and acceleration at t = 5?
Hint
The velocity is zero but the stipulated acceleration continues.
Worked solution
v = 0 and a = −2 m/s². The particle is momentarily at rest, not stationary over an interval.
03 / Add both parts of the path
Subtract positions to measure the return leg.
Watch: coordinate decreases while total distance increases
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 · After eight seconds
Find displacement and distance after 8 s.
Hint
The turn occurs inside the interval.
Worked solution
Final displacement = 80 − 64 = 16 m. The outward path is 25 m; return path is 25 − 16 = 9 m. Total distance = 34 m.
04 · Signed velocity
Find velocity at t = 8 and explain its sign.
Hint
v = 10 − 2t.
Worked solution
v = −6 m/s, meaning leftward motion. The particle is still to the right of the start, at x = 16 m.
04 / Crossing the origin is not the turnaround
The particle can continue left beyond its start.
05 · Return to start
When does the particle next reach x = 0, and what distance has it travelled then?
Hint
Solve 10t − t² = 0 and exclude departure.
Worked solution
The later root is t = 10 s. Distance = 25 + 25 = 50 m, despite displacement being zero.
06 · Beyond the start
Find position and distance at t = 12.
Hint
The coordinate is now negative.
Worked solution
Position = 120 − 144 = −24 m. Return leg = 25 − (−24) = 49 m. Total distance = 25 + 49 = 74 m.
05 / A question may start after departure
Use the turn only if it lies inside the requested interval.
07 · Interval crossing the turn
Find displacement and distance from t = 2 to t = 8.
Hint
Both endpoints have coordinate 16 m; the intermediate turn is at25 m.
Worked solution
Displacement = 16 − 16 = 0 m. Distance = (25 − 16) + (25 − 16) = 18 m.
08 · Entirely after the turn
Find displacement and distance from t = 6 to t = 8.
Hint
Positions are 24 m and16 m and velocity is negative throughout.
Worked solution
Displacement = 16 − 24 = −8 m. Distance = 8 m; no further split is needed.
09 · Average quantities
Find average speed and average velocity over the first 8 s.
Hint
Use total distance34 m and displacement16 m.
Worked solution
Average speed = 34/8 = 4.25 m/s. Average velocity = 16/8 = +2 m/s.
06 / A stopping model has a different continuation
Braking is normally specified only until rest.
10 · Vehicle stops
A vehicle has u = 10 m/s and a = −2 m/s² until it stops, then remains at rest. Find position and distance at t = 8.
Hint
The acceleration phase ends at5 s.
Worked solution
It stops at25 m after5 s, then stays there for3 s. Position = 25 m and distance = 25 m. The continued-acceleration particle model would give different answers.
11 · Stopping formula
Derive stopping distance for initial forward speed u > 0 and constant acceleration a < 0 until rest.
Hint
Set v = 0 in v² = u² + 2as.
Worked solution
s = −u²/(2a), which is positive because a is negative. With positive braking magnitude b = −a, this is u²/(2b).
07 / A squared equation loses the sign of velocity
Use the time or motion description to identify the event.
12 · Two velocities at a position
For u = 8 m/s and a = −2 m/s² continuing, find possible velocities when displacement is12 m.
Hint
v² = 64 − 48.
Worked solution
v = ±4 m/s. The particle reaches12 m once before turning and once after turning.
13 · Distinguish the visits
Find the corresponding times for question12.
Hint
Use v = 8 − 2t for each signed value.
Worked solution
v = +4 gives t = 2 s; v = −4 gives t = 6 s. Both are possible if the acceleration model continues through its turn at4 s.
14 · Absolute displacement shortcut
Is total distance always the absolute value of final displacement?
Hint
Look at a return journey.
Worked solution
Only when motion has no reversal within the interval. For the first8 s of the main model, |s| = 16 m but distance = 34 m.
08 / Split at reversals, not at the origin
Use the model to decide what happens after rest.
Find the zero of velocity and check whether it lies in the interval. Calculate endpoint and turning positions, then add the absolute position changes for distance. A stopping vehicle and a particle with continuing signed acceleration require different continuations.
Section 1 of 8 · Find whether velocity changes sign