| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Piecewise motion functions |
| Difficulty | Standard +0.3 This is a standard M1 kinematics question requiring integration of acceleration to find velocity, then integration to find displacement, followed by applying constant acceleration formulas. The piecewise nature adds mild complexity, but the techniques are routine and well-practiced. Slightly easier than average due to straightforward integration and arithmetic. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(1.8t^2/2\) \((+C)\) \((t = 0, v = 0)\) \(C = 0\) Expression is \(1.8t^2/2\) | M*1 B1 A1 | For using \(v = \int adt\) May be awarded in (ii). Accept c written and deleted. also for \(1.8t^2 +c\) |
| (ii) \(0.9t^3/3\) \((+K)\) | A1 | SR Award B1 for (s = 0, t = 0) K = 0 if not already given in (i), or +K included and limits used. |
| \(0.3 \times 64\) 19.2m AG \(u = 0.9 \times 4^2\) | M1 A1 D* M1 M1 | For using limits 0 to 4 (or equivalent) For using 'u' = v(4) For using \(s = ut + \frac{1}{2}x7.2t^2\) with non-zero u (s = 75.6) |
| \(s = 14.4 \times 3 + \frac{1}{2}7.2 \times 3^2\) 19.2 + 75.6 Displacement is 94.8m OR \(v = \int 7.2dt\) \(t = 0, v = 14.4, c = 14.4\) \(s = \int 7.2t + 14.4dt\) \(t = 0, s = 0, k = 0\) | M1 A1 A1 M1 | For adding distances for the two distinct stages For finding v(4) Integration and finding non-zero integration constant Nb Using t=4, v=14.4 gives c = -14.4 \(s = \int 7.2t - 14.4dt\) Integration and finding integration constant. Nb t=4 with s=19.2 and v=7.2t-14.4 gives k=19.2 Substituting t = 3 (OR 7 into s = 3.6t^2 - 14.4t + 19.2) (s=75.6) (OR s = 3.6 x7^2 - 14.4x7 + 19.2) Adding two distinct stages OR \(s = 3.6 x7^2 - 14.4x7+19.2 =94.8\) final M1A1 |
| \(s=3.6x3^2+14.4x3\) 19.2 + 75.6 = 94.8 Displacement is 94.8m |
(i) $1.8t^2/2$ $(+C)$ $(t = 0, v = 0)$ $C = 0$ Expression is $1.8t^2/2$ | M*1 B1 A1 | For using $v = \int adt$ May be awarded in (ii). Accept c written and deleted. also for $1.8t^2 +c$
(ii) $0.9t^3/3$ $(+K)$ | A1 | SR Award B1 for (s = 0, t = 0) K = 0 if not already given in (i), or +K included and limits used.
$0.3 \times 64$ 19.2m AG $u = 0.9 \times 4^2$ | M1 A1 D* M1 M1 | For using limits 0 to 4 (or equivalent) For using 'u' = v(4) For using $s = ut + \frac{1}{2}x7.2t^2$ with non-zero u (s = 75.6)
$s = 14.4 \times 3 + \frac{1}{2}7.2 \times 3^2$ 19.2 + 75.6 Displacement is 94.8m OR $v = \int 7.2dt$ $t = 0, v = 14.4, c = 14.4$ $s = \int 7.2t + 14.4dt$ $t = 0, s = 0, k = 0$ | M1 A1 A1 M1 | For adding distances for the two distinct stages For finding v(4) Integration and finding non-zero integration constant Nb Using t=4, v=14.4 gives c = -14.4 $s = \int 7.2t - 14.4dt$ Integration and finding integration constant. Nb t=4 with s=19.2 and v=7.2t-14.4 gives k=19.2 Substituting t = 3 (OR 7 into s = 3.6t^2 - 14.4t + 19.2) (s=75.6) (OR s = 3.6 x7^2 - 14.4x7 + 19.2) Adding two distinct stages OR $s = 3.6 x7^2 - 14.4x7+19.2 =94.8$ final M1A1
$s=3.6x3^2+14.4x3$ 19.2 + 75.6 = 94.8 Displacement is 94.8m | | |
---A particle starts from rest at a point $A$ at time $t = 0$, where $t$ is in seconds. The particle moves in a straight line. For $0 \leq t \leq 4$ the acceleration is $1.8t \text{ m s}^{-2}$, and for $4 \leq t \leq 7$ the particle has constant acceleration $7.2 \text{ m s}^{-2}$.
\begin{enumerate}[label=(\roman*)]
\item Find an expression for the velocity of the particle in terms of $t$, valid for $0 \leq t \leq 4$. [3]
\item Show that the displacement of the particle from $A$ is 19.2 m when $t = 4$. [4]
\item Find the displacement of the particle from $A$ when $t = 7$. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2007 Q5 [12]}}