| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Definite integration in mechanics context |
| Difficulty | Standard +0.3 This is a straightforward mechanics integration question requiring students to integrate given velocity functions and interpret results. The piecewise function is clearly defined, integration of polynomials is routine, and the constant velocity section requires only multiplication. While it involves multiple steps and careful attention to intervals, it demands no novel insight—just systematic application of standard techniques (∫v dt for distance, total distance/time for average speed). Slightly above average difficulty due to the piecewise nature and multi-part structure, but still a standard textbook-style question. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([s = t^2/2 - 0.1t^3/3]\) | M1* | For integrating to find \(s\) for \(0 \leq t \leq 5\) |
| \([s_1 = 25/2 - 0.1 \times 125/3]\) | DM1* | For obtaining \(s_1\) by using limits 0 to 5 or having zero for constant of integration (can be implied) and substituting \(t = 5\) |
| \(s_1 = 8.33\) | A1 | 3 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(s_2 = 2.5 \times 40\) | M1, A1 | For using \(s = v(5) \times (45-5)\) for \(5 \leq t \leq 45\) |
| \([s = 9t^2/2 - 0.1t^3/3 - 200t \text{ for } 45 \leq t \leq 50]\) | M1 | For integrating to find \(s\) for \(45 \leq t \leq 50\) and implying the use of limits 45 and 50 or equivalent via constant of integration |
| \(s_3 = [9(50)^2/2 - 0.1(50)^3/3 - 200(50)] - [9(45)^2/2 - 0.1(45)^3/3 - 200(45)]\) | A1 | For applying the limits at 45 and 50 correctly or equivalent via constant of integration |
| \([= 8.33]\) |
## Question 5:
**Part (i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[s = t^2/2 - 0.1t^3/3]$ | M1* | For integrating to find $s$ for $0 \leq t \leq 5$ |
| $[s_1 = 25/2 - 0.1 \times 125/3]$ | DM1* | For obtaining $s_1$ by using limits 0 to 5 or having zero for constant of integration (can be implied) and substituting $t = 5$ |
| $s_1 = 8.33$ | A1 | **3 marks total** |
**Part (ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $s_2 = 2.5 \times 40$ | M1, A1 | For using $s = v(5) \times (45-5)$ for $5 \leq t \leq 45$ |
| $[s = 9t^2/2 - 0.1t^3/3 - 200t \text{ for } 45 \leq t \leq 50]$ | M1 | For integrating to find $s$ for $45 \leq t \leq 50$ and implying the use of limits 45 and 50 or equivalent via constant of integration |
| $s_3 = [9(50)^2/2 - 0.1(50)^3/3 - 200(50)] - [9(45)^2/2 - 0.1(45)^3/3 - 200(45)]$ | A1 | For applying the limits at 45 and 50 correctly or equivalent via constant of integration |
| $[= 8.33]$ | | |5 A particle $P$ moves in a straight line. $P$ starts from rest at $O$ and travels to $A$ where it comes to rest, taking 50 seconds. The speed of $P$ at time $t$ seconds after leaving $O$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, where $v$ is defined as follows.
$$\begin{aligned}
\text { For } 0 \leqslant t \leqslant 5 , & v = t - 0.1 t ^ { 2 } \\
\text { for } 5 \leqslant t \leqslant 45 , & v \text { is constant } \\
\text { for } 45 \leqslant t \leqslant 50 , & v = 9 t - 0.1 t ^ { 2 } - 200
\end{aligned}$$
(i) Find the distance travelled by $P$ in the first 5 seconds.\\
(ii) Find the total distance from $O$ to $A$, and deduce the average speed of $P$ for the whole journey from $O$ to $A$.
\hfill \mbox{\textit{CAIE M1 2013 Q5 [9]}}