A-Level Mathematics May/June 2025 Q7(a): A particle X moves along a straight track, starting from a point O at time t = 0. The d…

1 answer

• ▲0
  accepted ✓

  The displacement of the particle X is given by $s = 3t^{\frac{3}{2}} - 6t$.

  To find the velocity, $v$, we differentiate the displacement $s$ with respect to time $t$. $v = \frac{ds}{dt} = \frac{d}{dt}(3t^{\frac{3}{2}} - 6t)$ $v = 3 \times \frac{3}{2}t^{\frac{3}{2}-1} - 6$ $v = \frac{9}{2}t^{\frac{1}{2}} - 6$

  The particle is instantaneously at rest when its velocity is zero, i.e., $v=0$. $\frac{9}{2}t^{\frac{1}{2}} - 6 = 0$ $\frac{9}{2}t^{\frac{1}{2}} = 6$ $t^{\frac{1}{2}} = 6 \times \frac{2}{9} = \frac{12}{9} = \frac{4}{3}$ $t = (\frac{4}{3})^2 = \frac{16}{9}$ So, the particle is at rest at $t = \frac{16}{9}$ s.

  To find the total distance travelled between $t=0$ and $t=16$, we must consider the change in direction at $t = \frac{16}{9}$. We need to calculate the displacement at the start, the turning point, and the end of the interval.

  Displacement at $t=0$: $s(0) = 3(0)^{\frac{3}{2}} - 6(0) = 0$ m.

  Displacement at $t=\frac{16}{9}$: $s(\frac{16}{9}) = 3(\frac{16}{9})^{\frac{3}{2}} - 6(\frac{16}{9})$ $s(\frac{16}{9}) = 3 \times (\sqrt{\frac{16}{9}})^3 - \frac{32}{3} = 3 \times (\frac{4}{3})^3 - \frac{32}{3}$ $s(\frac{16}{9}) = 3 \times \frac{64}{27} - \frac{32}{3} = \frac{64}{9} - \frac{96}{9} = -\frac{32}{9}$ m.

  Displacement at $t=16$: $s(16) = 3(16)^{\frac{3}{2}} - 6(16)$ $s(16) = 3 \times (\sqrt{16})^3 - 96 = 3 \times 4^3 - 96$ $s(16) = 3 \times 64 - 96 = 192 - 96 = 96$ m.

  The total distance is the sum of the magnitudes of the displacements for the two parts of the journey: $t=0$ to $t=\frac{16}{9}$ and $t=\frac{16}{9}$ to $t=16$.

  Total distance = $|s(\frac{16}{9}) - s(0)| + |s(16) - s(\frac{16}{9})|$ Total distance = $|-\frac{32}{9} - 0| + |96 - (-\frac{32}{9})|$ Total distance = $\frac{32}{9} + |96 + \frac{32}{9}|$ Total distance = $\frac{32}{9} + \frac{864}{9} + \frac{32}{9} = \frac{928}{9}$

  Total distance = $\frac{928}{9}$ m, or approximately 103 m (3 s.f.).

  ---

  How the marks are awarded

  • M1 — Attempting to differentiate the displacement function $s(t)$. This is shown by the power of $t$ decreasing in at least one term, for example in the step $v = 3 \times \frac{3}{2}t^{\frac{1}{2}-1} - 6$.
  • A1 — Obtaining the correct expression for velocity, $v = \frac{9}{2}t^{\frac{1}{2}} - 6$.
  • M1 — Setting the velocity expression to zero and attempting to solve for $t$. This includes rearranging the equation and correctly squaring to find $t$, leading to $t = (\frac{4}{3})^2$.
  • DM1 — Using the calculated time of rest, $t=\frac{16}{9}$, to split the time interval and evaluating the displacement $s$ at $t=0$, $t=\frac{16}{9}$, and $t=16$.
  • DM1 — Correctly combining the displacements to find the total distance. This involves adding the magnitude of the displacement in the first interval to the magnitude of the displacement in the second interval: $|s(\frac{16}{9}) - s(0)| + |s(16) - s(\frac{16}{9})|$.
  • A1 — Obtaining the final correct answer of $\frac{928}{9}$ or an answer which rounds to 103.

  Common mistakes

  • Calculating the net displacement $s(16) - s(0) = 96$ m instead of the total distance travelled. This mistake ignores the fact that the particle changes direction.
  • Making an algebraic error when solving $v=0$, for example incorrectly squaring $\frac{4}{3}$ or making a sign error.
  • Incorrectly calculating the total distance from the displacements, for example by calculating $s(16) - s(\frac{16}{9})$ and forgetting the first part of the journey from $t=0$ to $t=\frac{16}{9}$.
  • Making an arithmetic error when evaluating $s(t)$ at fractional or large values of $t$, particularly with the power of $3/2$.

  Examiner tip: For total distance problems, always find when the velocity is zero to check for turning points within the time interval, as this is where the direction of motion changes.

  ---

  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

  @markscheme_answers

Your answer

Sign in to answer this question.