Additional Mathematics Synthesis Guide 35: Kinematics from Calculus — Displacement, Velocity, Acceleration, Direction Changes and Total Distance

s(t) → differentiate → v(t) → differentiate → a(t); reverse by integration, then use conditions to recover constants.

1. The kinematics chain

If s(t) is signed position or displacement from an origin, then

• v(t)=ds/dt
• a(t)=dv/dt=d²s/dt²

Integration reverses these relationships, but introduces constants determined by initial conditions.

2. Choose a positive direction

One-dimensional motion needs an orientation. If rightward is positive, leftward velocity is negative. If upward is positive, downward velocity is negative.

The sign convention should be established before interpreting velocity or displacement.

3. Velocity is signed; speed is not

If v=−4 m/s, the particle moves at speed 4 m/s in the negative direction.

Speed is |v|. Therefore two particles can have equal speed and opposite velocities.

4. Worked differentiation example

Let

s(t)=t³−6t²+9t+2.

Then

v(t)=3t²−12t+9=3(t−1)(t−3)

and

a(t)=6t−12.

The factorised velocity immediately exposes possible turning times t=1 and t=3.

5. A direction change requires a velocity sign change

At v=0, a particle is instantaneously at rest. It changes direction only if the sign of v changes through that time.

For v=3(t−1)(t−3):

• 0≤t<1: v>0;
• 1<t<3: v<0;
• t>3: v>0.

So direction changes at both t=1 and t=3.

6. Position at the turning times

Using s=t³−6t²+9t+2:

s(1)=6

and

s(3)=2.

The particle moves forward to position 6, reverses to position 2, then reverses again.

7. Displacement versus total distance

From t=0 to t=4:

s(0)=2, s(1)=6, s(3)=2, s(4)=6.

Net displacement is

s(4)−s(0)=4.

Total distance is

|6−2|+|2−6|+|6−2|=12.

Distance must be split at every direction change.

8. Why endpoint subtraction is not total distance

Endpoint subtraction records only net signed change. Any motion that is later undone cancels in displacement but still contributes to distance.

Total distance measures path length along the line, not final position change.

9. Acceleration and speeding up are different ideas

A positive acceleration does not automatically mean increasing speed.

• If v and a have the same sign, speed increases.
• If v and a have opposite signs, speed decreases.

For example, v<0 and a>0 means the negative velocity is moving toward zero, so the particle slows down.

10. Worked speed interpretation

At t=2 in the earlier example:

v(2)=−3

and

a(2)=0.

At that exact instant acceleration is zero. To decide whether speed is increasing immediately before or after, inspect signs of v and a on neighbouring intervals.

11. Reconstructing velocity from acceleration

Suppose

a(t)=6t−4

and v(0)=3.

Integrate:

v(t)=3t²−4t+C.

Using v(0)=3 gives C=3, so

v(t)=3t²−4t+3.

12. Reconstructing position from velocity

If v(t)=3t²−4t+3 and s(0)=5, integrate:

s(t)=t³−2t²+3t+C.

Using s(0)=5 gives C=5.

The integration constant stores the initial state removed by differentiation.

13. A constant acceleration example

If a=−2 m/s² and v(0)=10 m/s, then

v=10−2t.

The particle is at rest when t=5 s. If the model continues, velocity becomes negative after 5 s, so the particle reverses direction.

A zero velocity time is a candidate turning point; the post-zero sign confirms the reversal.

14. Velocity graph and displacement

Because v=ds/dt, the signed area under a velocity-time graph over [a,b] equals

s(b)−s(a).

Area below the time axis contributes negative displacement.

15. Velocity graph and total distance

Total distance is the area under the speed graph |v(t)|. Equivalently, split the velocity integral at every zero where direction changes and take absolute values of each signed displacement segment.

This is the kinematic version of the signed-area distinction in definite integration.

16. Acceleration graph and velocity change

Since a=dv/dt,

∫_t₁^t₂a(t)dt=v(t₂)−v(t₁).

A positive signed area under the acceleration graph means velocity has increased, though speed may or may not have increased.

17. Finding when a particle returns to a position

To find when a particle returns to its starting position, solve

s(t)=s(0).

One root is often t=0 itself. The later positive roots correspond to return times, provided they lie in the modelled time domain.

18. Maximum or minimum position

Position s(t) has a stationary point when v(t)=0. A sign change +→− in velocity gives a local maximum position; −→+ gives a local minimum position.

This is exactly the derivative-sign logic of Guide 31, with physical interpretation added.

19. Units are part of the answer

• position/displacement: m
• velocity: m/s
• acceleration: m/s²

If the time unit is minutes rather than seconds, every rate unit changes accordingly. A numerically correct derivative with the wrong unit is an incomplete physical interpretation.

20. Kinematic models have domains

A polynomial position function may be mathematically defined for all real t, but a physical question may specify t≥0 or 0≤t≤T.

Discard algebraic roots outside the time domain. Negative time may be mathematically valid but physically outside the stated model.

21. Common failure patterns

• Treating negative velocity as negative speed.
• Calling every v=0 time a direction change without checking velocity signs.
• Using |s(b)−s(a)| as total distance when the particle reverses.
• Assuming positive acceleration means increasing speed.
• Integrating acceleration or velocity and forgetting the constant determined by initial conditions.
• Ignoring the stated time domain.
• Forgetting units after differentiation or integration.
• Using a velocity integral as total distance without splitting where velocity changes sign.

22. A reliable kinematics routine

1. Record the time domain, origin and positive direction.
2. Identify which function is given: s, v or a.
3. Differentiate or integrate to obtain the required neighbouring state.
4. Use initial conditions to determine integration constants.
5. Factor velocity and find candidate rest/turning times.
6. Build a velocity sign chart to determine direction changes.
7. For total distance, split at every direction change.
8. Attach units and interpret the sign of the final result.

23. Practice set

1. If s=t²−4t+1, find v and a.
2. Find when the particle in Question 1 is at rest.
3. Does it change direction there?
4. If v=3t²−12t+9, find the rest times.
5. State the direction of motion on 0≤t<1, 1<t<3 and t>3.
6. If s=t³−6t²+9t+2, find s(0),s(1),s(3),s(4).
7. Find net displacement from t=0 to4.
8. Find total distance from t=0 to4.
9. If a=6t−4 and v(0)=3, find v.
10. If that v has s(0)=5, find s.
11. If a=−2 and v(0)=10, when is the particle at rest?
12. What happens to direction after that time if the model continues?
13. If v<0 and a>0, is speed necessarily increasing?
14. What does area under a velocity-time graph represent?
15. What does area under an acceleration-time graph represent?
16. How is total distance obtained from a velocity graph?
17. Why is v=0 only a candidate direction change?
18. What units does acceleration have if distance is metres and time seconds?
19. How do you find a later return to the starting position?
20. How do you identify a maximum position from velocity signs?

Answers

1. v=2t−4, a=2.
2. t=2.
3. Yes: v changes from negative to positive, so direction reverses.
4. 3(t−1)(t−3)=0 gives t=1,3.
5. Positive, negative, positive.
6. 2,6,2,6.
7. 6−2=4.
8. 4+4+4=12.
9. v=3t²−4t+3.
10. s=t³−2t²+3t+5.
11. t=5 s.
12. Velocity becomes negative, so direction reverses.
13. No. The opposite signs mean the magnitude of the negative velocity is decreasing locally.
14. Signed displacement over the interval.
15. Change in velocity.
16. Integrate |v|, or split signed displacement at every velocity sign change and add magnitudes.
17. Velocity may touch zero and keep the same sign.
18. m/s².
19. Solve s(t)=s(0) and retain later roots in the physical time domain.
20. Velocity changes from positive to negative.

24. What mastery looks like

Mastery means the learner moves fluently between position, velocity and acceleration, interprets signs physically, distinguishes rest from reversal, computes total distance piecewise, reconstructs functions from derivative data and uses units and time domains as part of the mathematics.

The transfer test is to give motion information in a different form — formula, graph or initial-value statement. If the learner can reconstruct the missing states and explain the actual motion without confusing displacement, velocity and distance, calculus has become a motion language rather than a formula chain.

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• Guide 33: Exponential and Logarithmic Functions — Laws, Change of Base, Inverse Graphs, Equations and Models
• Guide 34: Trigonometric Identity Engineering — Addition Formulae, Double Angles, Simplification, Proof and Equation Reduction
• Guide 36: Graph-to-Equation Reconstruction — Quadratics, Exponentials, Trigonometric Functions and Parameter Recovery
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