| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Velocity-time graph sketching |
| Difficulty | Moderate -0.8 This is a straightforward M1 kinematics question requiring students to sketch a velocity-time graph, use v = u + at for deceleration time, and calculate total time using area under the graph. All steps are routine applications of standard SUVAT formulas with no problem-solving insight required, making it easier than average. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| 1st section of graph (decreasing from 20) | B1 | |
| 2nd section of graph (constant then increasing) | B1 | |
| Figures 20, 8 and 25 marked | B1 (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(v = u + at \Rightarrow 8 = 20 - 0.4t\) | M1 | M1 for complete method producing equation in \(t\) only; allow \((20-8)/0.4\) |
| \(t = 30\) (s) | A1 (2) | A0 for \(t = -30\), even if changed to 30 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(1960 = (25 \times 20) + (30 \times 8) + (\frac{1}{2} \times 30 \times 12) + (60 \times 8) + 8t + \frac{1}{2} \times t \times 12\) | M1 A3 ft (2,1,0) | M1 for clear attempt to find whole area including at least one "\(\frac{1}{2}\)", equated to 1960 |
| \(1960 = 500 + 240 + 180 + 480 + 14t\) | DM1 A1 | DM1 dependent on first M1, for simplifying to collect \(t\) terms |
| \(T = 115 + 40 = 155\) | DM1 A1 (8) | Third DM1 for solving for \(T\); A1 for 155 |
## Question 4:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| 1st section of graph (decreasing from 20) | B1 | |
| 2nd section of graph (constant then increasing) | B1 | |
| Figures 20, 8 and 25 marked | B1 **(3)** | |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $v = u + at \Rightarrow 8 = 20 - 0.4t$ | M1 | M1 for complete method producing equation in $t$ only; allow $(20-8)/0.4$ |
| $t = 30$ (s) | A1 **(2)** | A0 for $t = -30$, even if changed to 30 |
### Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| $1960 = (25 \times 20) + (30 \times 8) + (\frac{1}{2} \times 30 \times 12) + (60 \times 8) + 8t + \frac{1}{2} \times t \times 12$ | M1 A3 ft (2,1,0) | M1 for clear attempt to find whole area including at least one "$\frac{1}{2}$", equated to 1960 |
| $1960 = 500 + 240 + 180 + 480 + 14t$ | DM1 A1 | DM1 dependent on first M1, for simplifying to collect $t$ terms |
| $T = 115 + 40 = 155$ | DM1 A1 **(8)** | Third DM1 for solving for $T$; A1 for 155 |
---\begin{enumerate}
\item A car is moving on a straight horizontal road. At time $t = 0$, the car is moving with speed $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and is at the point $A$. The car maintains the speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ for 25 s . The car then moves with constant deceleration $0.4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$, reducing its speed from $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The car then moves with constant speed $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ for 60 s . The car then moves with constant acceleration until it is moving with speed $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at the point $B$.\\
(a) Sketch a speed-time graph to represent the motion of the car from $A$ to $B$.\\
(b) Find the time for which the car is decelerating.
\end{enumerate}
Given that the distance from $A$ to $B$ is 1960 m ,\\
(c) find the time taken for the car to move from $A$ to $B$.
\hfill \mbox{\textit{Edexcel M1 2012 Q4 [13]}}