| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2020 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Total distance with direction changes |
| Difficulty | Standard +0.3 This is a straightforward mechanics question requiring students to (a) solve a cubic equation set to zero (which factors easily as t²(t-5)=0), and (b) integrate |v| over intervals where the particle changes direction. While it requires understanding of when to use absolute values for distance vs displacement, the algebra is routine and the conceptual demand is standard for AS-level mechanics. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = t^3 - 5t^2 = 0\); \(t^2(t-5)=0\) | M1 (AO 1.1a) | DR: Attempt to solve \(v=0\); allow M1 for at least one root found by substitution |
| So at rest when \(t=0\) or \(t=5\) s | A1 [2] (AO 1.1) | Both roots and no others |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| There is a change of direction when \(t=5\) | M1 (AO 3.4) | DR: Considering signed areas either side of \(t=5\) s |
| \(\int_0^5(t^3-5t^2)\,dt = \left[\frac{t^4}{4}-\frac{5t^2}{3}\right]_0^5 = -\frac{625}{12}\) | M1 (AO 3.4) | Algebraic integration seen attempted. SPECIAL CASE: \(\int_0^6(t^3-5t^2)\,dt = \left[\frac{t^4}{4}-\frac{5t^2}{3}\right]_0^6 = -36\); SCM1 for algebraic integration seen; SCA1 for \(-36\) m seen or distance 36 given |
| \(\int_5^6(t^3-5t^2)\,dt = \frac{193}{12}\) | ||
| Total distance \(\frac{625+193}{12} = \frac{409}{6} = 68.2\) m | A1 [3] (AO 1.1) | Correct to at least 2 sf |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = t^3 - 5t^2 = 0$; $t^2(t-5)=0$ | M1 (AO 1.1a) | DR: Attempt to solve $v=0$; allow M1 for at least one root found by substitution |
| So at rest when $t=0$ or $t=5$ s | A1 [2] (AO 1.1) | Both roots and no others |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| There is a change of direction when $t=5$ | M1 (AO 3.4) | DR: Considering signed areas either side of $t=5$ s |
| $\int_0^5(t^3-5t^2)\,dt = \left[\frac{t^4}{4}-\frac{5t^2}{3}\right]_0^5 = -\frac{625}{12}$ | M1 (AO 3.4) | Algebraic integration seen attempted. SPECIAL CASE: $\int_0^6(t^3-5t^2)\,dt = \left[\frac{t^4}{4}-\frac{5t^2}{3}\right]_0^6 = -36$; SCM1 for algebraic integration seen; SCA1 for $-36$ m seen or distance 36 given |
| $\int_5^6(t^3-5t^2)\,dt = \frac{193}{12}$ | | |
| Total distance $\frac{625+193}{12} = \frac{409}{6} = 68.2$ m | A1 [3] (AO 1.1) | Correct to at least 2 sf |
---6 In this question you must show detailed reasoning.\\
A particle moves in a straight line. Its velocity $v \mathrm {~ms} ^ { - 1 }$ after $t \mathrm {~s}$ is given by $\mathrm { v } = \mathrm { t } ^ { 3 } - 5 \mathrm { t } ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the times at which the particle is stationary.
\item Find the total distance travelled by the particle in the first 6 seconds.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2020 Q6 [5]}}