A-Level Mathematics October/November 2024 Q8: A particle P moves in a straight line, passing through a point O with velocity 4.2 ms¯¹…

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  The acceleration of the particle P is given by $a = 0.6t - 2.7$.

  First, we find the velocity, $v$, by integrating the acceleration with respect to time, $t$. $v = \int a \, dt = \int (0.6t - 2.7) \, dt$ $v = 0.3t^2 - 2.7t + C$

  We are given that when $t=0$, the velocity is $4.2$ ms⁻¹. We use this initial condition to find the constant of integration, $C$. $4.2 = 0.3(0)^2 - 2.7(0) + C$ $C = 4.2$

  So, the expression for velocity is: $v = 0.3t^2 - 2.7t + 4.2$

  The particle is at instantaneous rest when its velocity is zero. We set $v=0$ to find the times at which this occurs. $0.3t^2 - 2.7t + 4.2 = 0$ Multiplying by 10 to remove decimals: $3t^2 - 27t + 42 = 0$ Dividing by 3: $t^2 - 9t + 14 = 0$ Factorising the quadratic equation: $(t-2)(t-7) = 0$ So, the particle is at instantaneous rest at $t=2$ s and $t=7$ s.

  Next, we find the displacement, $s$, by integrating the velocity with respect to time. $s = \int v \, dt = \int (0.3t^2 - 2.7t + 4.2) \, dt$ $s = 0.1t^3 - 1.35t^2 + 4.2t + D$

  The particle passes through the point O at $t=0$, so we can set the displacement $s=0$ at $t=0$. This gives $D=0$. $s = 0.1t^3 - 1.35t^2 + 4.2t$

  We need to find the distance travelled between $t=2$ and $t=7$. We find the displacement at each of these times. At $t=2$: $s(2) = 0.1(2)^3 - 1.35(2)^2 + 4.2(2)$ $s(2) = 0.1(8) - 1.35(4) + 8.4$ $s(2) = 0.8 - 5.4 + 8.4 = 3.8$ m

  At $t=7$: $s(7) = 0.1(7)^3 - 1.35(7)^2 + 4.2(7)$ $s(7) = 0.1(343) - 1.35(49) + 29.4$ $s(7) = 34.3 - 66.15 + 29.4 = -2.45$ m

  The distance travelled between these two times is the magnitude of the change in displacement. Since the velocity is only zero at the start and end of the interval $[2, 7]$, the particle does not change direction in between. Distance = $|s(7) - s(2)| = |-2.45 - 3.8| = |-6.25|$ Distance = $6.25$ m.

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  How the marks are awarded

  • M1 — Awarded for the attempt to integrate the acceleration expression $a = 0.6t - 2.7$ to find velocity, shown by the line $v = 0.3t^2 - 2.7t + C$.
  • DM1 — Awarded for using the initial condition ($v=4.2$ at $t=0$) to find the constant $C=4.2$ and setting the resulting three-term quadratic expression for velocity equal to zero: $0.3t^2 - 2.7t + 4.2 = 0$.
  • A1 — Awarded for correctly solving the quadratic equation to find both times of instantaneous rest, $t=2$ and $t=7$.
  • DM1 — Awarded for the attempt to integrate the three-term velocity expression $v = 0.3t^2 - 2.7t + 4.2$ to find an expression for displacement, $s$.
  • A1 — Awarded for the correct expression for displacement, $s = 0.1t^3 - 1.35t^2 + 4.2t$. The constant of integration may be shown as $+C$ or correctly identified as zero.
  • M1 — Awarded for substituting both found values of $t$ (2 and 7) into the cubic displacement expression and calculating the difference between the resulting displacements, $s(7) - s(2)$ or $s(2) - s(7)$.
  • A1 — Awarded for the final correct answer of 6.25 m. The magnitude of the displacement change is required.

  Common mistakes

  • Forgetting the constant of integration, especially when finding the velocity function. If the constant $C=4.2$ is missed, the times for instantaneous rest will be incorrect, losing most of the marks.
  • Confusing displacement with distance. A candidate might calculate $s(7)$ or $s(2)$ and give one of those as the final answer, instead of finding the difference $|s(7) - s(2)|$.
  • Making algebraic errors when solving the quadratic equation for time, for example, by mis-factoring or incorrectly applying the quadratic formula.
  • Making arithmetic errors when substituting the values of $t$ into the cubic expression for displacement, which is easy to do with the decimals and powers involved.

  Examiner tip: This question rewards the systematic application of calculus to kinematics: remember that acceleration, velocity, and displacement are linked by integration, and initial conditions are crucial for finding the constants of integration.

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