| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2003 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Two vehicles: overtaking or meeting (graph-based) |
| Difficulty | Moderate -0.3 This is a standard M1 velocity-time graph question requiring basic SUVAT application and area interpretation. Students must find time from displacement using s=½at², then calculate distances at a specific instant. The problem is straightforward with clear given values and no conceptual surprises, making it slightly easier than average A-level. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks |
|---|---|
| Region under \(v = 2t\) from \(t = 0\) to \(t = T\) indicated | B1 |
| Answer | Marks |
|---|---|
| For attempting to set up and solve an equation using area \(\Delta = 16\) or for using \(s = \frac{1}{2}2t^2\) | M1 |
| For \(16 = \frac{1}{2}2T^2\) | A1 |
| \(T = 4\) | A1 |
| Answer | Marks |
|---|---|
| For using distance \(= 10 \times ans\text{(ii)}\) or for using the idea that the distance is represented by the area of the relevant parallelogram or by the area of the trapezium (with parallel sides 9 and 4 and height 10) minus the area of the triangle (with base 5 and height 10) | M1 |
| Distance is 40 m | A1ft |
# Question 3:
## Part (i)
| Region under $v = 2t$ from $t = 0$ to $t = T$ indicated | B1 | |
## Part (ii)
| For attempting to set up and solve an equation using area $\Delta = 16$ or for using $s = \frac{1}{2}2t^2$ | M1 | |
| For $16 = \frac{1}{2}2T^2$ | A1 | |
| $T = 4$ | A1 | |
**SR** (for candidates who find the height of the $\Delta$ but do not score M1) (Max 1 out of 3):
For $h/T = 2$ or $h = 2T$ or $v = 8$ B1
## Part (iii)
| For using distance $= 10 \times ans\text{(ii)}$ or for using the idea that the distance is represented by the area of the relevant parallelogram or by the area of the trapezium (with parallel sides 9 and 4 and height 10) minus the area of the triangle (with base 5 and height 10) | M1 | |
| Distance is 40 m | A1ft | |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{cb04a09c-af23-4e9d-b3da-da9e351fe879-2_556_974_1548_587}
The diagram shows the velocity-time graphs for the motion of two cyclists $P$ and $Q$, who travel in the same direction along a straight path. Both cyclists start from rest at the same point $O$ and both accelerate at $2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ up to a speed of $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Both then continue at a constant speed of $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. $Q$ starts his journey $T$ seconds after $P$.\\
(i) Show in a sketch of the diagram the region whose area represents the displacement of $P$, from $O$, at the instant when $Q$ starts.
Given that $P$ has travelled 16 m at the instant when $Q$ starts, find\\
(ii) the value of $T$,\\
(iii) the distance between $P$ and $Q$ when $Q$ 's speed reaches $10 \mathrm {~ms} ^ { - 1 }$.
\hfill \mbox{\textit{CAIE M1 2003 Q3 [6]}}