| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Finding constants from motion conditions |
| Difficulty | Standard +0.3 This is a straightforward variable acceleration question requiring standard calculus techniques: integrating velocity to find displacement, differentiating displacement to find velocity, and differentiating again for acceleration. Part (i) involves integration and solving a cubic equation. Part (ii) requires forming two simultaneous equations from given conditions. Part (iii) is routine differentiation. All steps are mechanical applications of standard methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| 7(i) | 0.6t2–0.12t3 = 0 | M1 |
| (t=0 or) t = 5 | A1 | |
| ∫vdt=0.2t3 – 0.03t4 | *M1 | For integrating the velocity |
| OP=[0.2 × 53 – 0.03 × 54]–[0] | DM1 | Use limits to find OP |
| Distance= 6.25 m | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| 7(ii) | k×53+c × 55 = 6.25 | B1 |
| v=3kt2+ 5ct4 | *M1 | For differentiating s to find v |
| 1.25=3k × 52 + 5c × 54 | DM1 | For using the given value of v=1.25 in the expression for v |
| Answer | Marks | Guidance |
|---|---|---|
| 75k + 3125c = 1.25 | M1 | For attempting to solve a pair of simultaneous equations in k and |
| Answer | Marks |
|---|---|
| k=0.1, c = –0.002 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 7(iii) | a=0.6t– 0.04t3 | M1 |
| At t=5, a = –2 Acceleration=–2ms–2 | A1 |
Question 7:
--- 7(i) ---
7(i) | 0.6t2–0.12t3 = 0 | M1 | For attempting to solve v=0
(t=0 or) t = 5 | A1
∫vdt=0.2t3 – 0.03t4 | *M1 | For integrating the velocity
OP=[0.2 × 53 – 0.03 × 54]–[0] | DM1 | Use limits to find OP
Distance= 6.25 m | A1 | AG
5
--- 7(ii) ---
7(ii) | k×53+c × 55 = 6.25 | B1 | Using s=6.25 at t=5 to set up equation in k and c
v=3kt2+ 5ct4 | *M1 | For differentiating s to find v
1.25=3k × 52 + 5c × 54 | DM1 | For using the given value of v=1.25 in the expression for v
125k+3125c = 6.25
75k + 3125c = 1.25 | M1 | For attempting to solve a pair of simultaneous equations in k and
c and finding a value of either k or c
k=0.1, c = –0.002 | A1
5
--- 7(iii) ---
7(iii) | a=0.6t– 0.04t3 | M1 | For differentiating their expression for v
At t=5, a = –2 Acceleration=–2ms–2 | A1
2A particle moves in a straight line, starting from rest at a point $O$, and comes to instantaneous rest at a point $P$. The velocity of the particle at time $t$ s after leaving $O$ is $v$ m s$^{-1}$, where
$$v = 0.6t^2 - 0.12t^3.$$
\begin{enumerate}[label=(\roman*)]
\item Show that the distance $OP$ is 6.25 m. [5]
\end{enumerate}
On another occasion, the particle also moves in the same straight line. On this occasion, the displacement of the particle at time $t$ s after leaving $O$ is $s$ m, where
$$s = kt^3 + ct^5.$$
It is given that the particle passes point $P$ with velocity 1.25 m s$^{-1}$ at time $t = 5$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the values of the constants $k$ and $c$. [5]
\end{enumerate}
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the acceleration of the particle at time $t = 5$. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2019 Q7 [12]}}