| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2004 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Displacement from velocity by integration |
| Difficulty | Moderate -0.3 This is a straightforward integration question requiring students to integrate a polynomial velocity function and apply an initial condition, then substitute values. The mathematics is routine (integrating powers of t, solving a quartic that factors nicely), requiring no novel insight beyond standard M1 techniques. Slightly easier than average due to the clean polynomial and straightforward parts. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(10t^2 - 0.25t^4\) \((+C)\) | M1, DM1 | For integrating \(v\); for including constant of integration and attempting to evaluate it |
| Expression is \(10t^2 - 0.25t^4 - 36\) | A1 (×3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Displacement is 60 m | A1 ft (×1) | Dependent on both M marks in (i); ft if there is not more than one error in \(s(t)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((t^2 - 36)(1 - 0.25t^2) = 0\) | M1 | For attempting to solve \(s = 0\) (depends on both method marks in (i)) or \(\int_0^t v\,dt = 36\) (but not \(-36\)) for \(t^2\) by factors or formula method |
| Roots of quadratic are 4, 36 | A1 | |
| \(t = 2, 6\) | A1 ft (×3) | ft only from 3 term quadratic in \(t^2\) |
# Question 5:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $10t^2 - 0.25t^4$ $(+C)$ | M1, DM1 | For integrating $v$; for including constant of integration and attempting to evaluate it |
| Expression is $10t^2 - 0.25t^4 - 36$ | A1 (×3) | |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Displacement is 60 m | A1 ft (×1) | Dependent on both M marks in (i); ft if there is not more than one error in $s(t)$ |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(t^2 - 36)(1 - 0.25t^2) = 0$ | M1 | For attempting to solve $s = 0$ (depends on both method marks in (i)) or $\int_0^t v\,dt = 36$ (but not $-36$) for $t^2$ by factors or formula method |
| Roots of quadratic are 4, 36 | A1 | |
| $t = 2, 6$ | A1 ft (×3) | ft only from 3 term quadratic in $t^2$ |
---5 A particle $P$ moves in a straight line that passes through the origin $O$. The velocity of $P$ at time $t$ seconds is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, where $v = 20 t - t ^ { 3 }$. At time $t = 0$ the particle is at rest at a point whose displacement from $O$ is - 36 m .\\
(i) Find an expression for the displacement of $P$ from $O$ in terms of $t$.\\
(ii) Find the displacement of $P$ from $O$ when $t = 4$.\\
(iii) Find the values of $t$ for which the particle is at $O$.
\hfill \mbox{\textit{CAIE M1 2004 Q5 [7]}}