| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Find acceleration from distances/times |
| Difficulty | Standard +0.3 This is a standard SUVAT problem requiring systematic application of kinematic equations with algebraic manipulation. While it involves multiple unknowns and requires careful setup (using the 1.5 distance ratio to find acceleration, then applying results to part ii), the techniques are routine for M1 students and the question provides clear structure with part (i) showing the acceleration value, reducing problem-solving demand. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(s_{PQ} = 20\times10 - 0.5a\times10^2\) or \(s_{QR} = 20\times10 + 0.5a\times10^2\) | M1 | For use of \(s = vt - \frac{1}{2}at^2\) or \(s = ut + \frac{1}{2}at^2\) OE; suvat to find \(PQ\) or \(QR\) |
| \(s = 200-50a\) and \(1.5s = 200 + 50a\) | A1 | OE |
| \(1.5(200-50a) = 200+50a \rightarrow 100 = 125a \rightarrow a = 0.8 \text{ ms}^{-2}\) | B1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Distance \(QS = 20\times20 + \frac{1}{2}\times0.8\times20^2\) | M1 | Using \(s = ut + \frac{1}{2}at^2\) |
| Distance \(= 560\) m | A1 | |
| Average speed between \(Q\) and \(S = \frac{560}{20} = 28 \text{ ms}^{-1}\) | B1 |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s_{PQ} = 20\times10 - 0.5a\times10^2$ **or** $s_{QR} = 20\times10 + 0.5a\times10^2$ | M1 | For use of $s = vt - \frac{1}{2}at^2$ or $s = ut + \frac{1}{2}at^2$ OE; suvat to find $PQ$ or $QR$ |
| $s = 200-50a$ **and** $1.5s = 200 + 50a$ | A1 | OE |
| $1.5(200-50a) = 200+50a \rightarrow 100 = 125a \rightarrow a = 0.8 \text{ ms}^{-2}$ | B1 | AG |
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Distance $QS = 20\times20 + \frac{1}{2}\times0.8\times20^2$ | M1 | Using $s = ut + \frac{1}{2}at^2$ |
| Distance $= 560$ m | A1 | |
| Average speed between $Q$ and $S = \frac{560}{20} = 28 \text{ ms}^{-1}$ | B1 | |4 A car travels along a straight road with constant acceleration. It passes through points $P , Q , R$ and $S$. The times taken for the car to travel from $P$ to $Q , Q$ to $R$ and $R$ to $S$ are each equal to 10 s . The distance $Q R$ is 1.5 times the distance $P Q$. At point $Q$ the speed of the car is $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Show that the acceleration of the car is $0.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(ii) Find the distance $Q S$ and hence find the average speed of the car between $Q$ and $S$.\\
\hfill \mbox{\textit{CAIE M1 2019 Q4 [6]}}