A-Level Mathematics May/June 2024 Q6(b)(ii): (ii) Find the distance travelled by the particle in the first 16s.

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  To find the distance travelled, we must integrate the velocity function, $v = 10t^{\frac{1}{2}} - 2t - 8$. From part (b)(i), we know the particle changes direction at $t=1$. Therefore, we must calculate the distance travelled in the intervals $0 \le t \le 1$ and $1 \le t \le 16$ separately and add the magnitudes.

  First, find the displacement function, $s(t)$, by integrating $v(t)$: $s(t) = \int (10t^{\frac{1}{2}} - 2t - 8) \, dt$ $s(t) = \frac{10t^{\frac{3}{2}}}{3/2} - \frac{2t^2}{2} - 8t + c$ $s(t) = \frac{20}{3}t^{\frac{3}{2}} - t^2 - 8t + c$

  Assuming the particle starts at the origin, $s(0)=0$, so $c=0$.

  Interval 1: $0 \le t \le 1$ Displacement from $t=0$ to $t=1$ is $s(1) - s(0)$. $s(1) = \frac{20}{3}(1)^{\frac{3}{2}} - (1)^2 - 8(1) = \frac{20}{3} - 1 - 8 = \frac{20}{3} - 9 = -\frac{7}{3}$ m. $s(0) = 0$. Distance travelled, $d_1 = |s(1) - s(0)| = |-\frac{7}{3}| = \frac{7}{3}$ m.

  Interval 2: $1 \le t \le 16$ Displacement from $t=1$ to $t=16$ is $s(16) - s(1)$. $s(16) = \frac{20}{3}(16)^{\frac{3}{2}} - (16)^2 - 8(16)$ $s(16) = \frac{20}{3}(64) - 256 - 128 = \frac{1280}{3} - 384 = \frac{1280 - 1152}{3} = \frac{128}{3}$ m. Distance travelled, $d_2 = |s(16) - s(1)| = |\frac{128}{3} - (-\frac{7}{3})| = |\frac{128+7}{3}| = |\frac{135}{3}| = 45$ m.

  Total Distance Total distance travelled = $d_1 + d_2$ Total distance = $\frac{7}{3} + 45 = \frac{7}{3} + \frac{135}{3} = \frac{142}{3}$ m.

  Total distance = $47.3$ m (3 s.f.)

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  How the marks are awarded

  • M1 — Awarded for attempting to integrate the velocity function, demonstrated by increasing the power of 't' in at least one term, for example, $t^{1/2}$ becoming $t^{3/2}$.
  • A1 — Awarded for the correct integrated expression for displacement: $s = \frac{20}{3}t^{\frac{3}{2}} - t^2 - 8t$. The constant of integration 'c' is not required.
  • DM1 — Awarded for substituting the limits for one of the time intervals (either $t=0$ and $t=1$, or $t=1$ and $t=16$) into the integrated displacement function. This is dependent on the first M1 mark.
  • DM1 — Awarded for correctly evaluating the displacement for both intervals ($t=0$ to $t=1$ and $t=1$ to $t=16$) and adding their magnitudes. This is dependent on the first M1 mark.
  • A1 — Awarded for the final correct total distance of $\frac{142}{3}$ m. An equivalent decimal answer like 47.3 or 47.33 is also acceptable.

  Common mistakes

  • Calculating the net displacement from $t=0$ to $t=16$ by evaluating $s(16) - s(0)$, which ignores the change of direction at $t=1$. This gives an incorrect answer of $128/3$ m.
  • Making a sign error when calculating the displacement in the second interval, for example calculating $s(16) - s(1)$ as $\frac{128}{3} - \frac{7}{3}$ instead of $\frac{128}{3} - (-\frac{7}{3})$.
  • Incorrectly integrating the term $10t^{1/2}$, for example by forgetting to divide by the new power of $3/2$, leading to an incorrect displacement function.
  • Finding the displacements for the two intervals correctly but then subtracting them or failing to take the absolute value of the negative displacement, showing a misunderstanding of the concept of total distance.

  Examiner tip: To find total distance travelled, always check for changes in direction (where v=0) and integrate over each time interval separately, then sum the absolute values of the displacements.

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