| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Acceleration from velocity differentiation |
| Difficulty | Standard +0.3 This question involves straightforward differentiation of a polynomial to find acceleration, solving a quadratic equation for when acceleration equals zero, and integration of a polynomial for displacement. All techniques are standard calculus applications with no conceptual challenges or problem-solving insight required beyond routine method application. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.08d Evaluate definite integrals: between limits3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a(t) = 0.00012t^2 - 0.012t + 0.288\) | M1* | For attempting to differentiate \(v(t)\) |
| \([a(t) = 0.00012(t^2 - 100t + 2400) = 0.00012(t-40)(t-60) = 0]\) | dM1* | For setting \(a(t) = 0\) and attempting to solve a three term quadratic |
| \(a(t) = 0\) when \(t = 40\) and \(t = 60\) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([0.00001t^4 - 0.002t^3 + 0.144t^2]\) | M1† | For attempting to integrate \(v(t)\) |
| \([0.00001(100)^4 - 0.002(100)^3 + 0.144(100)^2]\) | dM1† | Integration attempted using correct limits \(t = 0\) to \(t = 100\) |
| Displacement is \(440\text{ m}\) | A1 | |
| Total: 3 |
# Question 3:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a(t) = 0.00012t^2 - 0.012t + 0.288$ | M1* | For attempting to differentiate $v(t)$ |
| $[a(t) = 0.00012(t^2 - 100t + 2400) = 0.00012(t-40)(t-60) = 0]$ | dM1* | For setting $a(t) = 0$ and attempting to solve a three term quadratic |
| $a(t) = 0$ when $t = 40$ and $t = 60$ | A1 | |
| | **Total: 3** | |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[0.00001t^4 - 0.002t^3 + 0.144t^2]$ | M1† | For attempting to integrate $v(t)$ |
| $[0.00001(100)^4 - 0.002(100)^3 + 0.144(100)^2]$ | dM1† | Integration attempted using correct limits $t = 0$ to $t = 100$ |
| Displacement is $440\text{ m}$ | A1 | |
| | **Total: 3** | |
---3 A particle $P$ moves along a straight line for 100 s . It starts at a point $O$ and at time $t$ seconds after leaving $O$ the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, where
$$v = 0.00004 t ^ { 3 } - 0.006 t ^ { 2 } + 0.288 t$$
(i) Find the values of $t$ at which the acceleration of $P$ is zero.\\
(ii) Find the displacement of $P$ from $O$ when $t = 100$.
\hfill \mbox{\textit{CAIE M1 2015 Q3 [6]}}