| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Distance from velocity function using calculus |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question requiring routine differentiation for acceleration and integration for distance. Part (a) is basic graph sketching, part (b) is direct differentiation of a given function, and part (c) involves standard integration of polynomials across three intervals. All techniques are standard M1 content 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 area3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks |
|---|---|
| 6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 6(a) | Correct for 0≤t≤5 | B1 |
| Correct for 5≤t≤7 | B1 | |
| Correct for 7≤t≤13.5 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 6(b) | a=−2t by differentiating | M1 |
| a=−12 | A1 |
| Answer | Marks |
|---|---|
| 6(c) | 5 6 7 13.5 |
| Answer | Marks |
|---|---|
| 0 5 6 7 | M1 |
| Answer | Marks |
|---|---|
| 0 5 6 7 | A1 |
| Answer | Marks |
|---|---|
| 3 | M1 |
| All correct | A1 |
| s=84.25 | A1 |
Question 6:
6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PPMMTT
9709/42 Cambridge International AS & A Level – Mark Scheme May/June 2020
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no “follow through” from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2020 Page 5 of 11
Question | Answer | Marks
--- 6(a) ---
6(a) | Correct for 0≤t≤5 | B1
Correct for 5≤t≤7 | B1
Correct for 7≤t≤13.5 | B1
3
--- 6(b) ---
6(b) | a=−2t by differentiating | M1
a=−12 | A1
2
--- 6(c) ---
6(c) | 5 6 7 13.5
s=(2t+1)dt+(36−t2)dt+ (36−t2)dt+ (2t−27)dt
0 5 6 7 | M1
5 6 7 13.5
s=(2t+1)dt+(36−t2)dt+ (36−t2)dt+ (2t−27)dt
0 5 6 7 | A1
t3
s=[t2 +t]+[36t− ]+t2 −27t
3 | M1
All correct | A1
s=84.25 | A1
5A particle $P$ moves in a straight line. The velocity $v \text{ ms}^{-1}$ at time $t$ s is given by
$$v = 2t + 1 \quad \text{for } 0 \leqslant t \leqslant 5,$$
$$v = 36 - t^2 \quad \text{for } 5 \leqslant t \leqslant 7,$$
$$v = 2t - 27 \quad \text{for } 7 \leqslant t \leqslant 13.5.$$
\begin{enumerate}[label=(\alph*)]
\item Sketch the velocity-time graph for $0 \leqslant t \leqslant 13.5$. [3]
\item Find the acceleration at the instant when $t = 6$. [2]
\item Find the total distance travelled by $P$ in the interval $0 \leqslant t \leqslant 13.5$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2020 Q6 [10]}}