A-Level Mathematics May/June 2025 Q6(b): It is given that in the interval 0 ≤ t ≤ 3 the velocity of P is always positive. Find t…

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  The velocity of the particle P is given by $v(t) = (2t+1)^{\frac{3}{2}} - 2t^2$. The distance of P from its initial position A is the displacement, $s(t)$, which is found by integrating the velocity function with respect to time. Since the velocity is always positive for $0 \le t \le 3$, the distance travelled is equal to the displacement.

  From part (a), the maximum velocity occurs at $t = 1.5$ s.

  To find the distance, we calculate the definite integral of $v(t)$ from $t=0$ to $t=1.5$.

  $s = \int_{0}^{1.5} \left( (2t+1)^{\frac{3}{2}} - 2t^2 \right) \, dt$

  First, find the indefinite integral: $\int \left( (2t+1)^{\frac{3}{2}} - 2t^2 \right) \, dt = \frac{(2t+1)^{\frac{5}{2}}}{\frac{5}{2} \times 2} - \frac{2t^3}{3} + c$ $= \frac{1}{5}(2t+1)^{\frac{5}{2}} - \frac{2}{3}t^3 + c$

  Now, evaluate the definite integral: $s = \left[ \frac{1}{5}(2t+1)^{\frac{5}{2}} - \frac{2}{3}t^3 \right]_{0}^{1.5}$

  Substitute the upper limit, $t=1.5$: $\left( \frac{1}{5}(2(1.5)+1)^{\frac{5}{2}} - \frac{2}{3}(1.5)^3 \right) = \left( \frac{1}{5}(4)^{\frac{5}{2}} - \frac{2}{3}(3.375) \right)$ $= \left( \frac{1}{5}(32) - 2.25 \right) = \frac{32}{5} - \frac{9}{4} = 6.4 - 2.25 = 4.15$

  Substitute the lower limit, $t=0$: $\left( \frac{1}{5}(2(0)+1)^{\frac{5}{2}} - \frac{2}{3}(0)^3 \right) = \left( \frac{1}{5}(1)^{\frac{5}{2}} - 0 \right) = \frac{1}{5} = 0.2$

  Calculate the distance: Distance = (Value at $t=1.5$) - (Value at $t=0$) Distance = $4.15 - 0.2 = 3.95$

  As a fraction: $s = \frac{32}{5} - \frac{9}{4} - \frac{1}{5} = \frac{31}{5} - \frac{9}{4} = \frac{124 - 45}{20} = \frac{79}{20}$

  The distance of P from A at this instant is $3.95$ m.

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  How the marks are awarded

  • M1 — Awarded for attempting to integrate the velocity function. This is shown by increasing the power of at least one term, e.g., $(2t+1)^{3/2}$ becomes a form of $(2t+1)^{5/2}$ or $-2t^2$ becomes a form of $t^3$.
  • A1 — Awarded for the correct integrated expression, $\frac{1}{5}(2t+1)^{\frac{5}{2}} - \frac{2}{3}t^3$. The constant of integration is not required.
  • DM1 — Awarded for substituting the limits $t=1.5$ (the time for maximum velocity found in part (a)) and $t=0$ into their integrated expression and subtracting the results. This is dependent on the first M1 mark.
  • A1 — Awarded for the correct final answer of $3.95$ or $\frac{79}{20}$, derived from correct working.

  Common mistakes

  • Forgetting to divide by the derivative of the inner function (the '2' from '2t') when integrating the term $(2t+1)^{3/2}$.
  • Incorrectly evaluating the definite integral by only calculating the value at $t=1.5$ and assuming the value at $t=0$ is zero. In this case, $\int v(t) dt$ is not zero at $t=0$.
  • Confusing distance with velocity and simply substituting $t=1.5$ into the velocity function $v(t)$ instead of integrating it.
  • Making arithmetic errors when substituting fractional or decimal values, especially with powers, for example miscalculating $(4)^{5/2}$ as $10$ instead of $32$.

  Examiner tip: Remember that distance is the definite integral of speed (the magnitude of velocity), and displacement is the definite integral of velocity; if velocity is always positive, these two quantities are the same.

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