June 2025 Paper 3 Q9
9
In this question you must show detailed reasoning.
A particle moves in a straight line. Its velocity \(t\) seconds after leaving a fixed point on the line is \(v\) m s−1 where \(v = 2 + 3t + 0.6t^2\).
Find the distance travelled by the particle from time \(t = 0\) until the instant when its acceleration is 15 m s−2. [6]
| Scheme | Marks | AO |
|---|---|---|
| DR | ||
| \((a =)\ 3 + 1.2t\) | B1 | 2.1 |
| \(3 + 1.2t = 15\) | M1* | 1.1 |
| M1* | 1.1 | |
| \((s =)\ 2t + \frac{3}{2}t^2 + \frac{0.6}{3}t^3\ (+c)\) | A1 | 1.1 |
| Distance travelled \(= \left[2t + \frac{3}{2}t^2 + \frac{0.6}{3}t^3\right]_0^{10} = 2(10) + \frac{3}{2}(10)^2 + \frac{0.6}{3}(10)^3\) | M1dep* | 3.4 |
| 370 (m) | B1 | 2.1 |
| [6] |
Notes
B1: Correct expression for the acceleration of the particle
B0 if \(s = 3 + 1.2t\)
M1*: Sets their linear expression for \(a\) equal to 15
For reference \(t = 10\)
M1*: Attempt to integrate expression – at least three terms with correct \(t^2\) or \(t^3\) term
A1: Correct expression for the displacement (allow un-simplified)
A0 if \(a = \int v\,\mathrm{d}t\)
M1dep*: Correct use of limits with their positive value of \(t\) – must see as a minimum all three values (either simplified or un-simplified), for example, if correct then simplified is \(20 + 150 + 200\). No consideration of lower limit of zero is required but if considered then must be correct, for example, \(2(10) + \frac{3}{2}(10)^2 + \frac{0.6}{3}(10)^3 - 2(0) + \frac{3}{2}(0)^2 + \frac{0.6}{3}(0)^3\) (so incorrect bracketing) is M0 (as DR)
M0 for \(20 + 150 + 200 + c\) unless \(c = 0\) is explicitly stated (so not just implied by seeing 370 as a final answer)
A correct answer of 370 does not imply this mark
This M mark is dependent on both previous M marks
B1: No justification that \(v \gt 0\) for all values of \(t\) is required for full marks. Ignore any working seen for this mark but full marks can only be awarded for a completely correct solution