| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Read and interpret velocity-time graph |
| Difficulty | Standard +0.3 This is a straightforward M1 mechanics question requiring basic calculus (differentiation to find acceleration) and interpretation of velocity-time graphs (finding distance as area under graph). Part (ii)(b) involves algebraic manipulation to show a given result, and part (c) is a simple deduction from completing the square. All techniques are standard and the question guides students through each step with no novel problem-solving required. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)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 |
|---|---|---|
| \([2 \times \frac{1}{2}(1 + 9) \cdot 400]\) | M1 | For using area property for distance |
| Approximation is \(4000 \text{ m}\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | For using the gradient property for acceleration | |
| Accelerations are \(0.02 \text{ m s}^{-2}\) and \(-0.02 \text{ m s}^{-2}\) | A1 | 2; Accept deceleration is \(0.02 \text{ m s}^{-2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | For using \(a = dv/dt\) and attempting to solve \(a = 0.02\) or \(a = -0.02\). | |
| \(0.04 - 0.0001t = \pm 0.02\) | A1 ft | |
| Values of \(t\) are \(200\) and \(600\) | A1 | 3 |
| Answer | Marks |
|---|---|
| \(v_1 - v = 0.02t + 1 - 0.04t + 0.00005t^2\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 0.00005(t - 200)^2 - 1\) | B1 | 2; AG |
| Answer | Marks |
|---|---|
| \(t = 200 \rightarrow -1 \leq v_1 - v\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| when \(t = 400 \rightarrow v_1 - v \leq 1\) | B1 | 2 |
**(i) (a)**
$[2 \times \frac{1}{2}(1 + 9) \cdot 400]$ | M1 | For using area property for distance
Approximation is $4000 \text{ m}$ | A1 | 2
**(i) (b)**
| M1 | For using the gradient property for acceleration
Accelerations are $0.02 \text{ m s}^{-2}$ and $-0.02 \text{ m s}^{-2}$ | A1 | 2; Accept deceleration is $0.02 \text{ m s}^{-2}$
**(ii) (a)**
| M1 | For using $a = dv/dt$ and attempting to solve $a = 0.02$ or $a = -0.02$.
$0.04 - 0.0001t = \pm 0.02$ | A1 ft |
Values of $t$ are $200$ and $600$ | A1 | 3
**(ii) (b)**
$v_1 - v = 0.02t + 1 - 0.04t + 0.00005t^2$ | B1 |
$v_1 - v = [0.00005t^2 - 0.02t + 2 - 1]$
$= 0.00005(t^2 - 400t + 40000) - 1$
$= 0.00005(t - 200)^2 - 1$ | B1 | 2; AG
**(ii) (c)**
For using $(v_1 - v)_{\min}$ occurs when
$t = 200 \rightarrow -1 \leq v_1 - v$ | B1 |
For using $(v_1 - v)_{\max}$ occurs when $t = 0$ and
when $t = 400 \rightarrow v_1 - v \leq 1$ | B1 | 27 A tractor travels in a straight line from a point $A$ to a point $B$. The velocity of the tractor is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at time $t \mathrm {~s}$ after leaving $A$.\\
(i)\\
\includegraphics[max width=\textwidth, alt={}, center]{2bd9f770-65b1-48c2-bf58-24e732bb6988-4_668_1091_397_568}
The diagram shows an approximate velocity-time graph for the motion of the tractor. The graph consists of two straight line segments. Use the graph to find an approximation for
\begin{enumerate}[label=(\alph*)]
\item the distance $A B$,
\item the acceleration of the tractor for $0 < t < 400$ and for $400 < t < 800$.\\
(ii) The actual velocity of the tractor is given by $v = 0.04 t - 0.00005 t ^ { 2 }$ for $0 \leqslant t \leqslant 800$.\\
(a) Find the values of $t$ for which the actual acceleration of the tractor is given correctly by the approximate velocity-time graph in part (i).
For the interval $0 \leqslant t \leqslant 400$, the approximate velocity of the tractor in part (i) is denoted by $v _ { 1 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(b) Express $v _ { 1 }$ in terms of $t$ and hence show that $v _ { 1 } - v = 0.00005 ( t - 200 ) ^ { 2 } - 1$.
\item Deduce that $- 1 \leqslant v _ { 1 } - v \leqslant 1$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2011 Q7 [11]}}