| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2010 |
| Session | June |
| Marks | 13 |
| Topic | Variable acceleration (1D) |
| Type | Piecewise motion functions |
| Difficulty | Standard +0.3 This is a standard mechanics question requiring integration of piecewise acceleration functions, verification of boundary conditions, sketching a velocity-time graph, and calculating distance. While it involves multiple steps and careful handling of the piecewise function, all techniques are routine A-level mechanics procedures with no novel problem-solving required. The 13 marks reflect length rather than conceptual difficulty. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Attempt to obtain \(v(t)\) by integration: For first stage \(v = 30t - 3t^2 + (4)\) (constant of integration) | M1 A1 | |
| Show \(A = 0\), using \(v(0) = 0\) | A1 | |
| Calculate \(v(10) = 0\) | B1 | |
| For second stage \(v = 3t^2 - 90t + (B)\) | A1 | |
| Show \(B = 600\) using \(v(10) = 0\) | A1 | |
| Calculate \(v(20) = 1200 - 1800 + 600 = 0\) | A1 | [7] |
| (b) Sketch of \(v(t)\): Inverted parabola above axis for \(0 \leq t \leq 10\); Upright parabola below axis for \(10 \leq t \leq 20\) | B1 B1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Distance travelled obtained by integrating the modulus of each of the velocity formulae and adding: \(\int_0^{10}(30t - 3t^2)dt - \int_{10}^{20}(3t^2 - 90t + 600)dt\) | M1 M1 | |
| \([15t^2 - t^3]_0^{10} - [t^3 - 45t^2 + 600t]_{10}^{20}\) | A1 | |
| \(= 500 - (-500) = 1000\) m | A1 | [4] |
### (i)
**(a)** Attempt to obtain $v(t)$ by integration: For first stage $v = 30t - 3t^2 + (4)$ (constant of integration) | M1 A1 |
Show $A = 0$, using $v(0) = 0$ | A1 |
Calculate $v(10) = 0$ | B1 |
For second stage $v = 3t^2 - 90t + (B)$ | A1 |
Show $B = 600$ using $v(10) = 0$ | A1 |
Calculate $v(20) = 1200 - 1800 + 600 = 0$ | A1 | [7]
**(b)** Sketch of $v(t)$: Inverted parabola above axis for $0 \leq t \leq 10$; Upright parabola below axis for $10 \leq t \leq 20$ | B1 B1 | [2]
### (ii)
Distance travelled obtained by integrating the modulus of each of the velocity formulae and adding: $\int_0^{10}(30t - 3t^2)dt - \int_{10}^{20}(3t^2 - 90t + 600)dt$ | M1 M1 |
$[15t^2 - t^3]_0^{10} - [t^3 - 45t^2 + 600t]_{10}^{20}$ | A1 |
$= 500 - (-500) = 1000$ m | A1 | [4]A particle moves along a straight line under the action of a variable force. The acceleration is given by
$$a = \begin{cases}
30 - 6t, & \text{for } 0 \leqslant t \leqslant 10 \\
6t - 90, & \text{for } 10 \leqslant t \leqslant 20
\end{cases}$$
where time $t$ is measured in seconds and $a$ in m s$^{-2}$. The particle is at rest at the origin at $t = 0$.
\begin{enumerate}[label=(\roman*)]
\item \begin{enumerate}[label=(\alph*)]
\item Find the velocity $v$ of the particle in terms of $t$. Verify that $v = 0$ when $t = 10$ and $t = 20$. [7]
\item Sketch the velocity-time graph for the motion. [2]
\end{enumerate}
\item Calculate the total distance travelled by the particle. [4]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2010 Q12 [13]}}