A-Level Mathematics May/June 2025 Q6(a): A particle P moves in a straight line and passes through the point A at time t = 0. The…

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  To find the maximum velocity, we first need to find the time at which the velocity is stationary. This occurs when the acceleration, $a = \frac{dv}{dt}$, is zero. The velocity is given by:

  $v = (2t+1)^{\frac{3}{2}} - 2t^2$

  First, we differentiate v with respect to t to find the acceleration, a:

  $a = \frac{dv}{dt} = \frac{d}{dt}((2t+1)^{\frac{3}{2}} - 2t^2)$

  Using the chain rule for the first term and the power rule for the second term:

  $\frac{dv}{dt} = \frac{3}{2}(2t+1)^{(\frac{3}{2}-1)} \cdot 2 - 2(2t)$

  $\frac{dv}{dt} = 3(2t+1)^{\frac{1}{2}} - 4t$

  For maximum or minimum velocity, we set $\frac{dv}{dt} = 0$:

  $3(2t+1)^{\frac{1}{2}} - 4t = 0$

  $3(2t+1)^{\frac{1}{2}} = 4t$

  To solve for t, we square both sides of the equation:

  $(3(2t+1)^{\frac{1}{2}})^2 = (4t)^2$

  $9(2t+1) = 16t^2$

  $18t + 9 = 16t^2$

  Rearranging into a standard quadratic form ($ax^2+bx+c=0$):

  $16t^2 - 18t - 9 = 0$

  We can solve this quadratic equation for t using the quadratic formula, $t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:

  $t = \frac{-(-18) \pm \sqrt{(-18)^2 - 4(16)(-9)}}{2(16)}$

  $t = \frac{18 \pm \sqrt{324 + 576}}{32}$

  $t = \frac{18 \pm \sqrt{900}}{32}$

  $t = \frac{18 \pm 30}{32}$

  This gives two possible values for t:

  $t = \frac{18+30}{32} = \frac{48}{32} = 1.5$

  $t = \frac{18-30}{32} = \frac{-12}{32} = -0.375$

  Since the time interval is $0 \le t \le 3$, we discard the negative solution. The stationary point occurs at $t = 1.5$ s.

  To find the maximum velocity, we substitute this value of t back into the original velocity equation. (We should also check the endpoints of the interval, $t=0$ and $t=3$, but the stationary point is the only candidate for a local maximum).

  At $t=1.5$:

  $v = (2(1.5)+1)^{\frac{3}{2}} - 2(1.5)^2$

  $v = (3+1)^{\frac{3}{2}} - 2(2.25)$

  $v = 4^{\frac{3}{2}} - 4.5$

  $v = (\sqrt{4})^3 - 4.5$

  $v = 2^3 - 4.5$

  $v = 8 - 4.5 = 3.5$

  Checking endpoints: At $t=0$, $v = (1)^{3/2} - 0 = 1$ ms⁻¹. At $t=3$, $v = (7)^{3/2} - 2(3)^2 = 7\sqrt{7} - 18 \approx 18.52 - 18 = 0.52$ ms⁻¹.

  Comparing the values (1, 3.5, 0.52), the maximum velocity is 3.5 ms⁻¹.

  Maximum velocity = 3.5 ms⁻¹

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  How the marks are awarded

  • M1 — Attempting to differentiate the velocity function $v = (2t+1)^{\frac{3}{2}} - 2t^2$. This is shown by correctly applying the chain rule to the first term or the power rule to the second term.
  • A1 — Obtaining the correct derivative, $\frac{dv}{dt} = 3(2t+1)^{\frac{1}{2}} - 4t$. The expression can be un-simplified.
  • DM1 — Setting the derivative equal to zero and correctly manipulating the equation to eliminate the square root by squaring both sides, leading to the equation $9(2t+1) = 16t^2$.
  • DM1 — Correctly solving the resulting three-term quadratic equation, $16t^2 - 18t - 9 = 0$, to find a positive value for t, which is $t=1.5$.
  • A1 — Substituting the valid value of $t=1.5$ back into the original velocity equation to find the final, correct maximum velocity of 3.5 ms⁻¹.

  Common mistakes

  • Incorrectly applying the chain rule when differentiating $(2t+1)^{\frac{3}{2}}$, often by forgetting to multiply by the derivative of the inner function (2).
  • Making an algebraic error when squaring the equation $3(2t+1)^{\frac{1}{2}} = 4t$, for example by forgetting to square the 3, resulting in an incorrect quadratic.
  • Successfully finding the correct value of $t=1.5$ but then stopping, failing to substitute it back into the original velocity equation to find the actual velocity value.
  • Finding the stationary point but failing to consider the endpoints of the interval ($t=0$ and $t=3$) to confirm that the stationary point gives the absolute maximum velocity within the given domain.

  Examiner tip: To find the absolute maximum or minimum of a function on a closed interval, always find the stationary points by setting the derivative to zero, and then evaluate the function at these points and at the interval's endpoints.

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