| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Multi-stage motion with velocity-time graph given |
| Difficulty | Moderate -0.8 This is a straightforward velocity-time graph question requiring basic kinematics: calculating area under the graph (trapezium rule) and using distance = area. All steps are routine M1 techniques with no problem-solving insight needed—simpler than average A-level maths questions. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks |
|---|---|
| 3 | (a) Distance = ½ x 4 x 9 + 16 x 9 or ½ (20 + 16) x 9 M1 |
Question 3:
3 | (a) Distance = ½ x 4 x 9 + 16 x 9 or ½ (20 + 16) x 9 M1
= 162 m A1
(2)
(b) Distance over last 5 s = ½(9 + u) x 5 M1
162 + ½(9 + u) x 5 = 200 M1 A1√
⇒ u = 6.2 m s–1 A1
(4)
(c) 6.2 = 9 + 5a M1 A1√
a = (–) 0.56 m s–2 A1
(3)\includegraphics{figure_2}
A sprinter runs a race of 200 m. Her total time for running the race is 25 s. Figure 2 is a sketch of the speed-time graph for the motion of the sprinter. She starts from rest and accelerates uniformly to a speed of 9 m s$^{-1}$ in 4 s. The speed of 9 m s$^{-1}$ is maintained for 16 s and she then decelerates uniformly to a speed of $u$ m s$^{-1}$ at the end of the race. Calculate
\begin{enumerate}[label=(\alph*)]
\item the distance covered by the sprinter in the first 20 s of the race, [2]
\item the value of $u$, [4]
\item the deceleration of the sprinter in the last 5 s of the race. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2005 Q3 [9]}}