| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Find distance from SUVAT |
| Difficulty | Moderate -0.8 This is a straightforward kinematics problem using standard SUVAT equations. Part (a) requires direct application of v = u + at with all values given. Part (b) needs s = ut + ½at² to find AC, then subtraction to get BC. Both parts are routine calculations with no problem-solving insight required, making this easier than average but not trivial due to the two-step nature of part (b). |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks |
|---|---|
| 1 | (a) ‘v = u + at’: 74 = 2 + a x 20 ⇒ a = 3.6 m s–2 |
| Answer | Marks |
|---|---|
| Hence BC = 1200 – 760 = 440 m | M1 A1 |
Question 1:
1 | (a) ‘v = u + at’: 74 = 2 + a x 20 ⇒ a = 3.6 m s–2
(b) ‘v2 = u2 + 2as’: 742 = 22 + 2 x 3.6 x AC
or ‘s = ut + ½at2’: AC = 2 x 20 + ½ x 3.6 x 202
⇒ AC = 760 m
Hence BC = 1200 – 760 = 440 m | M1 A1
(2)
M1 A1√
A1
B1√
(4)In taking off, an aircraft moves on a straight runway $AB$ of length 1.2 km. The aircraft moves from $A$ with initial speed $2 \text{ m s}^{-1}$. It moves with constant acceleration and 20 s later it leaves the runway at $C$ with speed $74 \text{ m s}^{-1}$. Find
\begin{enumerate}[label=(\alph*)]
\item the acceleration of the aircraft, [2]
\item the distance $BC$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2005 Q1 [6]}}