| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Total distance with direction changes |
| Difficulty | Moderate -0.3 This is a standard kinematics question with non-constant acceleration requiring differentiation to find velocity and acceleration, then optimization and root-finding. While it has many parts (7 sub-questions), each individual step uses routine A-level techniques: direct differentiation, solving quadratics, finding stationary points, and calculating distances. The multi-part structure and need to track sign changes elevates it slightly above pure routine, but it remains a straightforward application of calculus to mechanics without requiring novel insight. |
| Spec | 1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| 10 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(v = 36 + 6t - 6t^2\) | M1, A1 | Attempt at differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a = 6 - 12t\) | M1, F1 | Attempt at differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Take \(a = 0\) so \(t = 0.5\) | M1, A1 | Allow table if maximum indicated or implied. FT their \(a\). cao |
| and \(v = 37.5\) so \(37.5\) m s\(^{-1}\) | A1 | Accept no justification given that this is maximum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Either: Solving \(36 + 6t - 6t^2 = 0\) | M1 | A method for two roots using their \(v\) |
| so \(t = -2\) or \(t = 3\) | B1, E1 | Factorisation or formula or … of their expression. Shown |
| Or: Sub values in expression for \(v\); both shown to be zero; a quadratic so the only roots | M1, E1, B1 | Allow just 1 substitution shown. Both shown. Must be a clear argument |
| \(x(-2) = -34\); \(x(3) = 91\) | B1, B1 | cao; cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\ | x(3) - x(0)\ | + \ |
| \(= \ | 91 - 10\ | + \ |
| \(= 98\) so 98 m | A1 | cao. SC 1 for \(s(4) - s(0) = 64\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| At the SP of \(v\): \(x(-2) = -34 < 0\) and \(x(3) = 91 > 0\); also \(x(-4) = 42 > 0\) and \(x(6) = -98 < 0\) | M1 | Or any other valid argument e.g. find all the zeros, sketch, consider sign changes. Must have some working. If only a sketch, must have correct shape |
| [sketch of cubic with 3 roots shown] | B1 | Doing appropriate calculations e.g. find all 3 zeros; sketch cubic reasonably (showing 3 roots); sign changes in range |
| so three times | B1 | 3 times seen |
# Question 8:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| 10 | B1 | |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $v = 36 + 6t - 6t^2$ | M1, A1 | Attempt at differentiation |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $a = 6 - 12t$ | M1, F1 | Attempt at differentiation |
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| Take $a = 0$ so $t = 0.5$ | M1, A1 | Allow table if maximum indicated or implied. FT their $a$. cao |
| and $v = 37.5$ so $37.5$ m s$^{-1}$ | A1 | Accept no justification given that this is maximum |
## Part (v)
| Answer | Mark | Guidance |
|--------|------|----------|
| **Either:** Solving $36 + 6t - 6t^2 = 0$ | M1 | A method for two roots using their $v$ |
| so $t = -2$ or $t = 3$ | B1, E1 | Factorisation or formula or … of their expression. Shown |
| **Or:** Sub values in expression for $v$; both shown to be zero; a quadratic so the only roots | M1, E1, B1 | Allow just 1 substitution shown. Both shown. Must be a clear argument |
| $x(-2) = -34$; $x(3) = 91$ | B1, B1 | cao; cao |
## Part (vi)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\|x(3) - x(0)\| + \|x(4) - x(3)\|$ | M1 | Considering two parts |
| $= \|91 - 10\| + \|74 - 91\|$ | A1 | Either correct |
| $= 98$ so 98 m | A1 | cao. SC 1 for $s(4) - s(0) = 64$ |
## Part (vii)
| Answer | Mark | Guidance |
|--------|------|----------|
| At the SP of $v$: $x(-2) = -34 < 0$ and $x(3) = 91 > 0$; also $x(-4) = 42 > 0$ and $x(6) = -98 < 0$ | M1 | Or any other valid argument e.g. find all the zeros, sketch, consider sign changes. Must have some working. If only a sketch, must have correct shape |
| [sketch of cubic with 3 roots shown] | B1 | Doing appropriate calculations e.g. find all 3 zeros; sketch cubic reasonably (showing 3 roots); sign changes in range |
| so three times | B1 | 3 times seen |8 The displacement, $x \mathrm {~m}$, from the origin O of a particle on the $x$-axis is given by
$$x = 10 + 36 t + 3 t ^ { 2 } - 2 t ^ { 3 }$$
where $t$ is the time in seconds and $- 4 \leqslant t \leqslant 6$.\\
(i) Write down the displacement of the particle when $t = 0$.\\
(ii) Find an expression in terms of $t$ for the velocity, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, of the particle.\\
(iii) Find an expression in terms of $t$ for the acceleration of the particle.\\
(iv) Find the maximum value of $v$ in the interval $- 4 \leqslant t \leqslant 6$.\\
(v) Show that $v = 0$ only when $t = - 2$ and when $t = 3$. Find the values of $x$ at these times.\\
(vi) Calculate the distance travelled by the particle from $t = 0$ to $t = 4$.\\
(vii) Determine how many times the particle passes through O in the interval $- 4 \leqslant t \leqslant 6$.
\hfill \mbox{\textit{OCR MEI M1 2008 Q8 [19]}}