| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Piecewise motion functions |
| Difficulty | Standard +0.3 This is a straightforward piecewise motion problem requiring standard calculus techniques: differentiate to find acceleration and verify continuity at t=5, then integrate both velocity functions to find total distance. While it involves two pieces and requires careful calculation, the methods are routine M1 techniques with no conceptual challenges or novel problem-solving required. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(a_1(t) = 1.44t - 0.288t^2, a_2(t) = 2.4 - 0.48t\) \([a_1 = 1.44x5 - 0.288x25, a_2 = 2.4 - 0.48x5]\) \(a_1 = a_2 (= 0)\) → no instantaneous change | M1, A1, M1, A1 | For using \(a(t) = \dot{v}(t)\) or For evaluating \(a_1(5)\) and \(a_2(5)\) |
| (ii) \(s_1 = 0.24t^3 - 0.024t^4, s_2 = 1.2t^2 - 0.08t^3\) \([\frac{1}{2}(0.24x5^3 - 0.024x5^4) - (0 - 0)] + [(1.2x10^2 - 0.08x10^3) - (1.2x5^2 - 0.08x5^3)]\) Distance is \(35 \text{ m}\) | M1, A1, M1, A1 | For using \(s = \int vdt\) or For using limits 0 to 5 and 5 to 10 or equivalent |
(i) $a_1(t) = 1.44t - 0.288t^2, a_2(t) = 2.4 - 0.48t$ $[a_1 = 1.44x5 - 0.288x25, a_2 = 2.4 - 0.48x5]$ $a_1 = a_2 (= 0)$ → no instantaneous change | M1, A1, M1, A1 | For using $a(t) = \dot{v}(t)$ or For evaluating $a_1(5)$ and $a_2(5)$ | [4]
(ii) $s_1 = 0.24t^3 - 0.024t^4, s_2 = 1.2t^2 - 0.08t^3$ $[\frac{1}{2}(0.24x5^3 - 0.024x5^4) - (0 - 0)] + [(1.2x10^2 - 0.08x10^3) - (1.2x5^2 - 0.08x5^3)]$ Distance is $35 \text{ m}$ | M1, A1, M1, A1 | For using $s = \int vdt$ or For using limits 0 to 5 and 5 to 10 or equivalent | [4]
---6 A particle travels along a straight line. It starts from rest at a point $A$ on the line and comes to rest again, 10 seconds later, at another point $B$ on the line. The velocity $t$ seconds after leaving $A$ is
$$\begin{array} { r l l }
0.72 t ^ { 2 } - 0.096 t ^ { 3 } & \text { for } & 0 \leqslant t \leqslant 5 \\
2.4 t - 0.24 t ^ { 2 } & \text { for } & 5 \leqslant t \leqslant 10
\end{array}$$
(i) Show that there is no instantaneous change in the acceleration of the particle when $t = 5$.\\
(ii) Find the distance $A B$.
\hfill \mbox{\textit{CAIE M1 2010 Q6 [8]}}