| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Constant acceleration with algebraic unknowns |
| Difficulty | Moderate -0.3 This is a standard kinematics problem using constant acceleration equations (suvat). Part (a) requires straightforward substitution into s = (u+v)t/2 or s = ut + ½at². Part (b) needs finding the midpoint distance then solving a quadratic, which is routine for M1. The multi-step nature and 6 marks for part (b) elevate it slightly above trivial, but it remains a textbook exercise requiring no novel insight. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks |
|---|---|
| \(240 = \frac{1}{2}(u + 34)\cdot 10\) | M1 A1 |
| \(u = 14\) | A1 |
| (3) |
| Answer | Marks |
|---|---|
| \(34 = 14 + 10a \Rightarrow a = 2\) | M1 A1 |
| \(120 = 14t + \frac{1}{2} \times 2 \times t^2\) | M1 A1 |
| \(t^2 + 14t - 120 = 0\) | |
| Solving, \(t = -20\) or 6 | DM1 |
| \(t = 6\) | A1 |
| Answer | Marks |
|---|---|
| \(34 = 14 + 10a \Rightarrow a = 2\) | M1 A1 |
| \(v^2 = 14^2 + 2 \times 2 \times 120 \Rightarrow v = 26\) | M1 A1 |
| AND \(26 = 14 + 2t\) | M1 A1 |
| \(t = 6\) | DM1 A1 |
| (6) | |
| [9] |
## Part (a)
$240 = \frac{1}{2}(u + 34)\cdot 10$ | M1 A1
$u = 14$ | A1
| (3)
## Part (b)
$34 = 14 + 10a \Rightarrow a = 2$ | M1 A1
$120 = 14t + \frac{1}{2} \times 2 \times t^2$ | M1 A1
$t^2 + 14t - 120 = 0$ |
Solving, $t = -20$ or 6 | DM1
$t = 6$ | A1
**OR**
$34 = 14 + 10a \Rightarrow a = 2$ | M1 A1
$v^2 = 14^2 + 2 \times 2 \times 120 \Rightarrow v = 26$ | M1 A1
AND $26 = 14 + 2t$ | M1 A1
$t = 6$ | DM1 A1
| (6)
| [9]
**Notes for Question 4:**
**Q4(a)**
- First M1 for a complete method to produce an equation in $u$ only.
- First A1 for a correct equation. ($u^2 - 48u + 476 = 0$ oe is possible).
- Second A1 for $u = 14$.
**EITHER**
**Q4(b)**
- First M1 for an equation in $a$ only. (M0 if $v = 34$ when $s = 120$ is used)
- First A1 for $a = 2$. (This may have been found in part (a))
- Second M1 for a 3-term quadratic equation in $t$ only, allow sign errors (must have found a value of $a$. (M0 if $v = 34$ when $s = 120$ is used)
- Second A1 for a correct equation.
- Third M1 dependent on previous M1 for solving for $t$.
- Third A1 for $t = 6$
**OR**
- First M1 for an equation in $a$ only.
- First A1 for $a = 2$. (This may have been found in part (a))
- Second M1 for a complete method to obtain an equation in $t$ only, allow sign errors. (must have found a value of $a$)
- Second A1 for a correct equation.
- Third M1 dependent on previous M1 for solving for $t$.
- Third A1 for $t = 6$
---A lorry is moving along a straight horizontal road with constant acceleration. The lorry passes a point $A$ with speed $u \text{ m s}^{-1}$, $(u < 34)$, and 10 seconds later passes a point $B$ with speed $34 \text{ m s}^{-1}$. Given that $AB = 240$ m, find
\begin{enumerate}[label=(\alph*)]
\item the value of $u$, [3]
\item the time taken for the lorry to move from $A$ to the mid-point of $AB$. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2013 Q4 [9]}}