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DSE Physics Recap – Kinematics | Tia (2026)

Kinematics

DSE Physics

Tia (2026)

For use on Sunday, 6 September 2026

01

Motion Concepts

Distance, Displacement and Velocity

A particle moves along a straight horizontal line. Positions to the right of the origin are taken as positive. During each time interval shown in the table, the particle moves uniformly without changing direction.

Time, t / s 0 2 5 9
Position, x / m −4 8 2 14
  1. Calculate the displacement of the particle during:
    1. the first 2.0 s;
    2. the interval from t = 2.0 s to t = 5.0 s;
    3. the complete 9.0 s journey.
  2. Calculate the total distance travelled during the complete journey.
  3. Calculate, for the complete journey:
    1. the average speed;
    2. the average velocity.
  4. Determine the velocity of the particle during each of the three time intervals.
  5. State the times at which the particle changes its direction of motion.
  6. Explain why the magnitude of the average velocity is smaller than the average speed for this journey.
  7. A student makes the following statement:
    “Whenever the acceleration of an object is negative, the object must be slowing down.”
    State whether the statement is correct and explain your answer.
15 marks
02

Graphical Analysis

Velocity and Acceleration

The velocity–time graph below shows the motion of an object along a straight line. Motion in one direction is taken as positive.

v / m s⁻¹ t / s 10 2 0 −6 0 4 7 9.5 11 Stage I Stage II Stage III

Velocity–time graph for the object. The graph is drawn to scale.

  1. Describe the motion of the object during:
    1. Stage I, from t = 0 s to t = 4.0 s;
    2. Stage II, from t = 4.0 s to t = 7.0 s;
    3. Stage III, from t = 7.0 s to t = 11.0 s.
  2. Calculate the acceleration during:
    1. Stage I;
    2. Stage II;
    3. Stage III.
  3. Determine the time at which the object changes its direction of motion.
  4. Calculate the displacement of the object during the complete 11.0 s journey.
  5. Calculate the total distance travelled during the complete journey.
  6. Calculate the average velocity of the object during the complete journey.
  7. At t = 9.0 s, the object has a positive velocity and a negative acceleration. State whether it is speeding up or slowing down.
  8. At t = 10.0 s, both the velocity and acceleration are negative. State whether it is speeding up or slowing down. Explain your answer.
18 marks
03

Equations of Motion

Multi-stage Motion of a Train

A train is travelling along a straight horizontal track at 8.0 m s⁻¹. It accelerates uniformly over a distance of 120 m until its speed becomes 20.0 m s⁻¹. It then travels at this constant speed for 15.0 s. Finally, it decelerates uniformly and comes to rest over a distance of 100 m.

  1. For the first stage, calculate:
    1. the acceleration of the train;
    2. the time taken for its speed to increase from 8.0 m s⁻¹ to 20.0 m s⁻¹.
  2. Determine the speed of the train after it has travelled 60.0 m from the beginning of the first stage.
  3. Calculate the distance travelled during the constant-speed stage.
  4. For the final stage, calculate:
    1. the acceleration of the train, including its sign;
    2. the time taken for the train to come to rest.
  5. Calculate the total distance travelled from the beginning of the first stage until the train comes to rest.
  6. Calculate the average speed of the train over the complete journey.
  7. A student calculates the average speed by taking the arithmetic mean of 8.0 m s⁻¹, 20.0 m s⁻¹ and 0 m s⁻¹. Explain why this method is incorrect.
17 marks
04

Equations of Motion

Vertical Motion of Two Particles

Particle A is projected vertically upwards from the ground with an initial speed of 24.0 m s⁻¹. At the same instant, particle B is released from rest from a balcony 45.0 m above the ground. The particles move along the same vertical line and pass each other without colliding. Air resistance is negligible. Take upward as positive and use g = 9.8 m s⁻².

Particle B u = 0 Particle A u = 24.0 m s⁻¹ 45.0 m Positive

The diagram is not drawn to scale.

  1. Write an equation for the height yA of particle A above the ground at time t.
  2. Write an equation for the height yB of particle B above the ground at time t.
  3. Calculate the time at which the two particles pass each other.
  4. Calculate the height above the ground at which the particles pass each other.
  5. Calculate the velocity of each particle at the instant when they pass each other. State the direction of each velocity.
  6. Hence, determine their relative speed at that instant.
  7. Calculate the greatest height reached by particle A above the ground.
  8. Determine whether particle A is moving upwards or downwards when it passes particle B. Support your answer with a calculation.
  9. Explain why the time at which the particles pass each other can be found without using the value of g.
20 marks

Question 1

    1. First 2.0 s:
      Displacement = 8 − (−4)
      = +12 m
    2. From t = 2.0 s to t = 5.0 s:
      Displacement = 2 − 8
      = −6 m
    3. Complete journey:
      Displacement = final position − initial position
      = 14 − (−4)
      = +18 m
  2. The distances travelled in the three intervals are 12 m, 6 m and 12 m.
    Total distance = 12 + 6 + 12
    = 30 m
    1. Average speed = total distance / total time
      = 30 / 9
      = 3.33 m s⁻¹
    2. Average velocity = total displacement / total time
      = 18 / 9
      = +2.00 m s⁻¹
  4. From t = 0 s to t = 2.0 s:
    v = 12 / 2 = +6.0 m s⁻¹
    From t = 2.0 s to t = 5.0 s:
    v = −6 / 3 = −2.0 m s⁻¹
    From t = 5.0 s to t = 9.0 s:
    v = (14 − 2) / 4 = +3.0 m s⁻¹
  5. The velocity changes from positive to negative at t = 2.0 s, and from negative to positive at t = 5.0 s. These are the times at which the particle changes direction.
  6. Average speed is based on the total distance of 30 m, while the magnitude of the average velocity is based on the displacement of only 18 m. Part of the journey is in the negative direction, so some of the particle’s displacement is cancelled, but the distance travelled is not cancelled.
  7. The statement is not always correct. Negative acceleration means that the acceleration acts in the negative direction. Whether the object speeds up or slows down depends on the directions of both its velocity and acceleration.
    • If velocity is positive and acceleration is negative, the object slows down.
    • If velocity is negative and acceleration is negative, the object speeds up because the magnitude of its velocity increases.
Key distinction: Distance and speed are scalars. Displacement, velocity and acceleration are vectors, so their directions or signs must be considered.

Question 2

    1. During Stage I, the object moves in the positive direction and accelerates uniformly from 2.0 m s⁻¹ to 10.0 m s⁻¹.
    2. During Stage II, the object moves in the positive direction at a constant velocity of 10.0 m s⁻¹.
    3. During Stage III, the velocity decreases uniformly from +10.0 m s⁻¹ to −6.0 m s⁻¹. The object first slows down, stops momentarily at t = 9.5 s, and then reverses direction and speeds up in the negative direction.
    1. Stage I:
      a = (10 − 2) / (4 − 0)
      a = +2.0 m s⁻²
    2. Stage II:
      Velocity is constant.
      a = 0 m s⁻²
    3. Stage III:
      a = (−6 − 10) / (11 − 7)
      = −16 / 4
      a = −4.0 m s⁻²
  3. During Stage III:
    v = u + at
    0 = 10 + (−4)Δt
    Δt = 2.5 s

    Time after start = 7.0 + 2.5
    t = 9.5 s
  4. Displacement is the signed area between the graph and the time axis.
    Stage I displacement = ½(2 + 10)(4) = 24 m

    Stage II displacement = 10(3) = 30 m

    Stage III displacement = ½(10 + (−6))(4) = 8 m

    Total displacement = 24 + 30 + 8
    = +62 m
  5. For total distance, the area below the time axis must be treated as positive.
    Distance from 0 s to 7.0 s = 24 + 30 = 54 m

    Distance from 7.0 s to 9.5 s = ½(2.5)(10) = 12.5 m

    Distance from 9.5 s to 11.0 s = ½(1.5)(6) = 4.5 m

    Total distance = 54 + 12.5 + 4.5
    = 71 m
  6. Average velocity = total displacement / total time
    = 62 / 11
    = +5.64 m s⁻¹
  7. At t = 9.0 s, velocity is positive but acceleration is negative. The velocity and acceleration act in opposite directions. Therefore, the object is slowing down.
  8. At t = 10.0 s, both velocity and acceleration are negative. They act in the same direction, so the magnitude of the negative velocity is increasing. Therefore, the object is speeding up in the negative direction.
Velocity–time graphs: The gradient gives acceleration. Signed area gives displacement. The total area without signs gives distance travelled.

Question 3

    1. Using v² = u² + 2as:
      20.0² = 8.0² + 2a(120)
      400 = 64 + 240a
      a = 336 / 240
      a = 1.40 m s⁻²
    2. Using v = u + at:
      20.0 = 8.0 + 1.40t
      t = 12.0 / 1.40
      t = 8.57 s
  2. After travelling 60.0 m, the train is still in the first uniformly accelerated stage.
    v² = u² + 2as
    v² = 8.0² + 2(1.40)(60.0)
    v² = 64 + 168 = 232
    v = √232
    v = 15.2 m s⁻¹
    The speed is not halfway between 8.0 m s⁻¹ and 20.0 m s⁻¹ because equal distances under constant acceleration do not correspond to equal changes in speed.
  3. Distance = speed × time
    = 20.0 × 15.0
    = 300 m
    1. Using v² = u² + 2as:
      0² = 20.0² + 2a(100)
      0 = 400 + 200a
      a = −2.00 m s⁻²
      The negative sign indicates that the acceleration acts opposite to the positive direction of motion.
    2. v = u + at
      0 = 20.0 + (−2.00)t
      t = 10.0 s
  5. Total distance = 120 + 300 + 100
    = 520 m
  6. Total time:
    Total time = 8.57 + 15.0 + 10.0
    = 33.57 s

    Average speed = 520 / 33.57
    = 15.5 m s⁻¹
  7. The arithmetic mean method is incorrect because the train does not spend equal amounts of time at the three listed speeds. In addition, the speed changes continuously during the acceleration and deceleration stages. Average speed must be calculated using total distance divided by total time.
The equations of uniformly accelerated motion may be applied only within a stage where the acceleration is constant. A multi-stage journey must be separated into suitable stages.

Question 4

  1. For particle A:
    yA = ut + ½at²
    yA = 24.0t + ½(−9.8)t²
    yA = 24.0t − 4.9t²
  2. Particle B begins at a height of 45.0 m with zero initial velocity.
    yB = 45.0 + 0t + ½(−9.8)t²
    yB = 45.0 − 4.9t²
  3. When the particles pass each other, their heights are equal:
    yA = yB
    24.0t − 4.9t² = 45.0 − 4.9t²
    24.0t = 45.0
    t = 1.875 s ≈ 1.88 s
  4. Substitute t = 1.875 s into the equation for particle A:
    y = 24.0(1.875) − 4.9(1.875)²
    y = 45.0 − 17.23
    y = 27.8 m
  5. For particle A:
    vA = u + at
    = 24.0 − 9.8(1.875)
    vA = +5.63 m s⁻¹
    Particle A is moving upwards.
    For particle B:
    vB = 0 − 9.8(1.875)
    vB = −18.4 m s⁻¹
    Particle B is moving downwards.
  6. The particles are moving in opposite directions.
    Relative speed = |vA − vB|
    = |5.625 − (−18.375)|
    = 24.0 m s⁻¹
  7. At the greatest height, the velocity of particle A is zero.
    v² = u² + 2as
    0² = 24.0² + 2(−9.8)s
    0 = 576 − 19.6s
    s = 576 / 19.6
    s = 29.4 m
  8. The time taken for particle A to reach its greatest height is:
    v = u + at
    0 = 24.0 − 9.8t
    t = 24.0 / 9.8
    t = 2.45 s
    The particles pass each other at 1.875 s, which is earlier than 2.45 s. Therefore, particle A has not yet reached its greatest height and is still moving upwards.
  9. Both particles have the same gravitational acceleration, −g. Therefore, the terms involving gravitational acceleration in their position equations are identical:
    24.0t − ½gt² = 45.0 − ½gt²
    The −½gt² terms cancel, leaving:
    24.0t = 45.0
    Thus, their relative acceleration is zero and their initial relative speed remains 24.0 m s⁻¹ until they pass each other.
When two objects are in free fall at the same location, they have the same gravitational acceleration. Their relative acceleration is therefore zero, even though each object is accelerating relative to the ground.
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