Higher November 2021 Paper 6 Q16
16 The following kinematics formulas may be used in this question.
\[v = u + at \qquad s = ut + \tfrac{1}{2}at^2 \qquad v^2 = u^2 + 2as\]The initial velocity of a particle is 20 m/s.
The acceleration of the particle is −8 m/s².
After \(t\) seconds, the particle has travelled 25 m.
(a) Show that \(4t^2 - 20t + 25 = 0\). [3]
(b) Solve \(4t^2 - 20t + 25 = 0\). [3]
(c) Show that the particle is stationary when it has travelled 25 m. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Subst into correct formula (may be implied) and partial simplification | 2 | B2 for \(25 = 20t - \frac{1}{2} \times 8 \times t^2\) oe or \(25 = 20t + (-4)t^2\) or B1 for subst eg \(25 = 20t + \frac{1}{2}(-8)t^2\) | Only accept \(25 = 20t - 4t^2\) if subst seen For B1 condone ambiguity caused by missing brackets |
| \(25 = 20t - 4t^2\) seen and correct completion to \(4t^2 - 20t + 25 = 0\) | 1dep | Dep on previous 2 marks | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 2.5 oe | 3 | M2 for \((2t - 5)(2t - 5)\) or M1 for any two factors that give two correct terms when expanded or for partial factorisation \(2t(2t - 5) - 5(2t - 5)\) OR M2 for [\(t =\)] \(\dfrac{20 \pm \sqrt{400 - 400}}{8}\) or better or M1 for [\(t =\)] \(\dfrac{-(-20) \pm \sqrt{(-20)^2 - 4 \times 4 \times 25}}{2 \times 4}\) with at most one error | eg a sign error, short fraction line, short root, but condone missing brackets |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Shows \(v = 0\) and concludes “stationary” | 3 | M1 for [\(v^2 =\)] \(20^2 + 2(-8)25\) or [\(v =\)] \(20 + (-8) \times\) their (b) A1 \(v = 0\) If 0 scored, instead award SC2 for \(v = 0\) and other values substituted into a relevant equation as a correct check or SC1 for \(v = 0\) | |