A-Level Mathematics May/June 2024 Q6(a): A particle moves in a straight line, starting from a point O. The velocity of the parti…

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  The velocity of the particle is given by $v = kt^{\frac{1}{2}} - 2t - 8$.

  The maximum velocity occurs when the acceleration, $a$, is zero. The acceleration is the rate of change of velocity, so we must find $\frac{dv}{dt}$.

  $a = \frac{dv}{dt} = \frac{d}{dt}(kt^{\frac{1}{2}} - 2t - 8)$ $a = \frac{1}{2}kt^{-\frac{1}{2}} - 2$

  For maximum velocity, set $a=0$: $\frac{1}{2}kt^{-\frac{1}{2}} - 2 = 0$ $\frac{1}{2}kt^{-\frac{1}{2}} = 2$ $kt^{-\frac{1}{2}} = 4$ $t^{-\frac{1}{2}} = \frac{4}{k}$ $t^{\frac{1}{2}} = \frac{k}{4}$

  This is the time at which the maximum velocity occurs. We are given that the maximum velocity is $4.5 \text{ ms}^{-1}$. We substitute $t^{\frac{1}{2}} = \frac{k}{4}$ into the expression for $v$ and set $v=4.5$.

  Note that if $t^{\frac{1}{2}} = \frac{k}{4}$, then $t = (\frac{k}{4})^2 = \frac{k^2}{16}$.

  $v_{max} = k(t^{\frac{1}{2}}) - 2(t) - 8 = 4.5$ $k(\frac{k}{4}) - 2(\frac{k^2}{16}) - 8 = 4.5$

  Now, we solve this equation for $k$. $\frac{k^2}{4} - \frac{2k^2}{16} - 8 = 4.5$ $\frac{k^2}{4} - \frac{k^2}{8} - 8 = 4.5$ $\frac{2k^2 - k^2}{8} - 8 = 4.5$ $\frac{k^2}{8} - 8 = 4.5$ $\frac{k^2}{8} = 12.5$ $k^2 = 12.5 \times 8$ $k^2 = 100$ $k = \pm 10$

  Since it is given that $k$ is a positive constant, we have $k=10$. (Shown)

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  How the marks are awarded

  • M1 — For attempting to differentiate the expression for $v$ to find the acceleration, $a$. This is shown by the power of $t$ decreasing from $\frac{1}{2}$ to $-\frac{1}{2}$ in the first term.
  • A1 — For the correct expression for acceleration, $a = \frac{1}{2}kt^{-\frac{1}{2}} - 2$. This may be unsimplified.
  • DM1 — For setting the acceleration $a=0$ and attempting to solve for $t$ or a function of $t$ (like $t^{\frac{1}{2}}$) in terms of $k$. In the model answer, this is achieved by finding $t^{\frac{1}{2}} = \frac{k}{4}$.
  • M1 — For substituting the expression for $t$ (or $t^{\frac{1}{2}}$) found from $a=0$ back into the velocity equation and setting it equal to the given maximum velocity, 4.5. This creates an equation solely in terms of $k$.
  • A1 — For correctly simplifying the equation and solving to show that $k=10$. As this is a 'show that' question, all intermediate working must be correct to be awarded this final mark.

  Common mistakes

  • Incorrectly differentiating $kt^{\frac{1}{2}}$, often forgetting the factor of $\frac{1}{2}$, leading to $a = kt^{-\frac{1}{2}} - 2$.
  • Making an algebraic error when solving $a=0$, for example rearranging $\frac{1}{2}kt^{-\frac{1}{2}} = 2$ to $t^{\frac{1}{2}} = \frac{k}{2}$ instead of $\frac{k}{4}$.
  • After finding $t^{\frac{1}{2}} = \frac{k}{4}$, making a mistake when substituting back into the velocity equation, for example incorrectly calculating the term $-2t$.
  • Successfully finding $k^2 = 100$ but failing to state why $k=10$ is chosen over $k=-10$ by referencing the condition that $k$ is a positive constant.

  Examiner tip: Remember that the maximum or minimum value of a quantity (like velocity) occurs when its rate of change with respect to a variable (like time) is zero.

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

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