| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Verifying given motion properties |
| Difficulty | Moderate -0.8 This is a straightforward calculus mechanics question requiring only routine differentiation of a polynomial twice and substitution of t=1.5. The 'show that' format makes it easier as students know the target answer. No problem-solving or insight needed—pure mechanical application of standard techniques. |
| Spec | 1.07i Differentiate x^n: for rational n and sums3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = \frac{d(t^4 - 2t^3 + 5)}{dt}\) | M1* | Differentiates displacement, one wrong term max, ignore \(+c\) |
| \(v = 4 \times 1.5^3 - 6 \times 1.5^2\) | D*M1 | Substitutes \(t = 1.5\) in \(v(t)\) OR solves \(4t^3 - 6t^2 = 0\) for a \(+\)ve root |
| \(v = 0\) AG | A1 | \(0+c\) is A0 unless \(c\) is discarded |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a = \frac{d(4t^3 - 6t^2)}{dt}\) | M1* | Differentiates velocity, one wrong term max, ignore \(+c\) |
| \(a(1.5) = 12 \times 1.5^2 - 12 \times 1.5\) | D*M1 | Substitutes \(t = 1.5\) in \(a(t)\) OR solves \(12t^2 - 12t = 9\) for a \(+\)ve root |
| \(a = 9\) m s\(^{-2}\) AG | A1 | \(9+c\) is A0 unless \(c\) is discarded |
| [3] |
## Question 2(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = \frac{d(t^4 - 2t^3 + 5)}{dt}$ | M1* | Differentiates displacement, one wrong term max, ignore $+c$ |
| $v = 4 \times 1.5^3 - 6 \times 1.5^2$ | D*M1 | Substitutes $t = 1.5$ in $v(t)$ OR solves $4t^3 - 6t^2 = 0$ for a $+$ve root |
| $v = 0$ AG | A1 | $0+c$ is A0 unless $c$ is discarded |
| **[3]** | | |
## Question 2(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = \frac{d(4t^3 - 6t^2)}{dt}$ | M1* | Differentiates velocity, one wrong term max, ignore $+c$ |
| $a(1.5) = 12 \times 1.5^2 - 12 \times 1.5$ | D*M1 | Substitutes $t = 1.5$ in $a(t)$ OR solves $12t^2 - 12t = 9$ for a $+$ve root |
| $a = 9$ m s$^{-2}$ AG | A1 | $9+c$ is A0 unless $c$ is discarded |
| **[3]** | | |2 A particle $P$ moves in a straight line. The displacement of $P$ from a fixed point on the line is $\left( t ^ { 4 } - 2 t ^ { 3 } + 5 \right) \mathrm { m }$, where $t$ is the time in seconds. Show that, when $t = 1.5$,\\
(i) $P$ is at instantaneous rest,\\
(ii) the acceleration of $P$ is $9 \mathrm {~ms} ^ { - 2 }$.
\hfill \mbox{\textit{OCR M1 2013 Q2 [6]}}