| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Displacement from velocity by integration |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question testing basic calculus skills: substituting t=0 for initial velocity, integrating a simple polynomial to find displacement, and differentiating to find acceleration then solving. All three parts are routine applications of standard techniques with no problem-solving insight required, making it easier than average. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3 \text{ ms}^{-1}\) | B1 | MR \((0.6t^3 + 3)\), award B1 here |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = \int 0.6t^2 + 3\ dt\) | M1* | Integrates \(v\); MR \((0.6t^3 + 3)\) |
| \(x = 0.6t^3/3 + 3t\ (+c)\) | A1 | Accept with/without \(+c\); \(0.6t^4/4 + 3t\) is A0 |
| Substitutes \(1.5\) in expression for \(x\) | D*M1 | Needs integration and 2 terms in \(t\) |
| \(x(1.5) = 5.175 \text{ m}\) | A1 | Only without \(+c\). Accept \(5.17\), \(5.18\); MR \(5.26\) only gets A1ft |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a = d(0.6t^2 + 3)/dt\) | M1* | Differentiates \(v\); MR \((0.6t^3 + 3)\) gives \(t = 1.82(57...)\) |
| \(6 = 2 \times 0.6t\) | D*M1 | Plus attempt to solve \(a(t) = 6\) |
| \(v(5) = 18 \text{ ms}^{-1}\) | A1 | \(v(1.8257...) = 6.65\) (3 sf) |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3 \text{ ms}^{-1}$ | B1 | MR $(0.6t^3 + 3)$, award B1 here |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = \int 0.6t^2 + 3\ dt$ | M1* | Integrates $v$; MR $(0.6t^3 + 3)$ |
| $x = 0.6t^3/3 + 3t\ (+c)$ | A1 | Accept with/without $+c$; $0.6t^4/4 + 3t$ is A0 |
| Substitutes $1.5$ in expression for $x$ | D*M1 | Needs integration and 2 terms in $t$ |
| $x(1.5) = 5.175 \text{ m}$ | A1 | Only without $+c$. Accept $5.17$, $5.18$; MR $5.26$ only gets A1ft |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = d(0.6t^2 + 3)/dt$ | M1* | Differentiates $v$; MR $(0.6t^3 + 3)$ gives $t = 1.82(57...)$ |
| $6 = 2 \times 0.6t$ | D*M1 | Plus attempt to solve $a(t) = 6$ |
| $v(5) = 18 \text{ ms}^{-1}$ | A1 | $v(1.8257...) = 6.65$ (3 sf) |
---3 A particle $P$ travels in a straight line. The velocity of $P$ at time $t$ seconds after it passes through a fixed point $A$ is given by $\left( 0.6 t ^ { 2 } + 3 \right) \mathrm { ms } ^ { - 1 }$. Find\\
(i) the velocity of $P$ when it passes through $A$,\\
(ii) the displacement of $P$ from $A$ when $t = 1.5$,\\
(iii) the velocity of $P$ when it has acceleration $6 \mathrm {~ms} ^ { - 2 }$.
\hfill \mbox{\textit{OCR M1 2014 Q3 [8]}}