| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Topic | Variable acceleration (1D) |
| Type | Variable acceleration with initial conditions |
| Difficulty | Standard +0.3 This is a straightforward kinematics problem requiring integration of acceleration to find velocity and displacement, with standard initial conditions. All three parts follow routine procedures: integrate with given initial condition, substitute values, and solve a quadratic equation. While it requires multiple steps and careful tracking of signs, it involves no novel insight or complex problem-solving beyond standard A-level mechanics techniques. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(v = \int (12 - 6t) dt\) | M1 | Set up integral for \(v\). |
| \(= 12t - 3t^2 (+c)\) | A1 | Correct integration. Condone omission of "c". |
| \(v = 0\) when \(t = 0 \therefore c = 0\) | M1 | "c" dealt with explicitly. |
| When \(t = 4\), \(v = 48 - 48 = 0\) ms\(^{-1}\) | A1 | c.a.o. from correctly integrated \(a\). Accept correct answer obtained from a definite integral. [4] |
| (ii) \(x = \int_0^4 (12t - 3t^2) dt\) | M1 | Correct integral of c's \(v\), including limits (which may appear or be dealt with later). |
| \(= (6t^2 - t^3) | _0^4\) | A1 |
| \(= (96 - 64) - (0) = 32\) m | A1 | c.a.o. following use of limits or explicit treatment of "c". [3] |
| (iii) When \(x = 0\), \(6t^2 - t^3 = 0\) | M1 | Equation for \(x = 0\). ft c's expression for \(x\) in (ii) only if obtained by integration. Condone omission of consideration of "c" (= 0). |
| \(t \neq 0 \therefore t = 6\) sec | A1 | Solved and non-zero solution chosen. |
| \(\therefore v = 72 - 108 = -36\) ms\(^{-1}\) | A1 | c.a.o. [3] |
**(i)** $v = \int (12 - 6t) dt$ | M1 | Set up integral for $v$.
$= 12t - 3t^2 (+c)$ | A1 | Correct integration. Condone omission of "c".
$v = 0$ when $t = 0 \therefore c = 0$ | M1 | "c" dealt with explicitly.
When $t = 4$, $v = 48 - 48 = 0$ ms$^{-1}$ | A1 | c.a.o. from correctly integrated $a$. Accept correct answer obtained from a definite integral. [4]
**(ii)** $x = \int_0^4 (12t - 3t^2) dt$ | M1 | Correct integral of c's $v$, including limits (which may appear or be dealt with later).
$= (6t^2 - t^3)|_0^4$ | A1 | Correct integration. ft c's $v$.
$= (96 - 64) - (0) = 32$ m | A1 | c.a.o. following use of limits or explicit treatment of "c". [3]
**(iii)** When $x = 0$, $6t^2 - t^3 = 0$ | M1 | Equation for $x = 0$. ft c's expression for $x$ in (ii) only if obtained by integration. Condone omission of consideration of "c" (= 0).
$t \neq 0 \therefore t = 6$ sec | A1 | Solved and non-zero solution chosen.
$\therefore v = 72 - 108 = -36$ ms$^{-1}$ | A1 | c.a.o. [3]A particle $P$ is free to move along a straight line $Ox$. It starts from rest at $O$ and after $t$ seconds its acceleration $a\,\mathrm{m}\,\mathrm{s}^{-2}$ is given by $a = 12 - 6t$.
\begin{enumerate}[label=(\roman*)]
\item Find an expression in terms of $t$ for its velocity $v\,\mathrm{m}\,\mathrm{s}^{-1}$. Hence find the velocity of $P$ when $t = 4$. [4]
\item Find the displacement of $P$ from $O$ when $t = 4$. [3]
\item Find the velocity of $P$ when it returns to $O$. [3]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2014 Q10 [10]}}