| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Distance from velocity function using calculus |
| Difficulty | Standard +0.3 This is a straightforward kinematics problem using standard SUVAT equations and basic integration. Part (i) requires simple application of v = u + at and related formulas. Part (ii) involves integrating a quadratic velocity function and using the constraint that distances are equal, which is routine for M1. The sketch requires understanding of linear vs quadratic velocity graphs. All techniques are standard with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.08d Evaluate definite integrals: between limits3.02d Constant acceleration: SUVAT formulae3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| \([2.2 - 1.8 + 0.004t]\) | M1 | For using \(v = u + at\) (or \(v^2 = u^2 + 2as\)) |
| Time taken is 100s | A1 | |
| Distance is 200 m (or Time taken is 100s) | A1ft | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(s = k(100t^2 - t^3/3) (+C)\) | M1 | For integrating v(t) to find s(t) |
| \([k(100x100^2 - 100^3/3) = 200]\) | A1 | |
| DM1 | For using s(0) = 0 (may be implied) and s(100) = 200 | |
| \(k = 0.0003\) | A1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Speed is 3 ms⁻¹ | B1ft | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | For straight line segment, v(t) +ve and increasing throughout (including at t = 0) | |
| M1 | For parabolic segment through origin, with +ve slope | |
| Parabolic segment has decreasing slope; sketches correct relative to each other (line crosses curve once) | A1 | 3 |
**(i)**
$[2.2 - 1.8 + 0.004t]$ | M1 | For using $v = u + at$ (or $v^2 = u^2 + 2as$)
Time taken is 100s | A1 |
Distance is 200 m (or Time taken is 100s) | A1ft | 3 | ft s = 2t or 1.8t + 0.002t² (or t = s/2)
**(ii)**
**(a)**
$s = k(100t^2 - t^3/3) (+C)$ | M1 | For integrating v(t) to find s(t)
$[k(100x100^2 - 100^3/3) = 200]$ | A1 |
| DM1 | For using s(0) = 0 (may be implied) and s(100) = 200
$k = 0.0003$ | A1 | 4
**(b)**
Speed is 3 ms⁻¹ | B1ft | 1 | ft candidate's t and/or k.
**(iii)**
| M1 | For straight line segment, v(t) +ve and increasing throughout (including at t = 0)
| M1 | For parabolic segment through origin, with +ve slope
| | |
Parabolic segment has decreasing slope; sketches correct relative to each other (line crosses curve once) | A1 | 3 | Depends on both M marks\begin{enumerate}[label=(\roman*)]
\item A man walks in a straight line from $A$ to $B$ with constant acceleration $0.004 \text{ m s}^{-2}$. His speed at $A$ is $1.8 \text{ m s}^{-1}$ and his speed at $B$ is $2.2 \text{ m s}^{-1}$. Find the time taken for the man to walk from $A$ to $B$, and find the distance $AB$. [3]
\item A woman cyclist leaves $A$ at the same instant as the man. She starts from rest and travels in a straight line to $B$, reaching $B$ at the same instant as the man. At time $t$ s after leaving $A$ the cyclist's speed is $k(200t - t^2) \text{ m s}^{-1}$, where $k$ is a constant. Find
\begin{enumerate}[label=(\alph*)]
\item the value of $k$, [4]
\item the cyclist's speed at $B$. [1]
\end{enumerate}
\item Sketch, using the same axes, the velocity-time graphs for the man's motion and the woman's motion from $A$ to $B$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2007 Q6 [11]}}