| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Multi-stage motion with velocity-time graph given |
| Difficulty | Standard +0.3 This is a standard SUVAT/kinematics question with straightforward application of formulas. Part (i) uses area under v-t graph equals distance (triangle area = 4m). Parts (ii)-(iv) apply basic SUVAT equations and trapezium area calculations. All steps are routine M1 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \([\frac{1}{2} \times 2.5(\text{speed}_{\max}) = 4]\) | M1 | For using area property for distance |
| Greatest speed is 3.2 ms⁻¹ | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Greatest speed is 3.2 ms⁻¹ from \(2 \times \frac{1}{2} \times 1.25(\text{speed}_{\max}) V = 4\) | B1 | |
| (ii) \([V = 3x2]\) | M1 | For using \(a = \frac{(V - 0)}{(4.5 - 2.5)}\) or \(V = 0 +\) at |
| \(V = 6\) | A1 | [2] |
| (iii) | M1 | For using area property for distance from \(t = 2.5\) to \(t = 14.5\) |
| \(\frac{1}{2} \times 6(12 + T) = 48\) or \(\frac{1}{2} \times 6x2 + 6T + \frac{1}{2} \times 6(10 - T) = 48\) or \(\frac{1}{2} \times 6x2 + 6(10 - \tau) + \frac{1}{2} \times 6\tau = 48\) | A1ft | |
| \(t = 8.5\) | A1 | [3] from 4.5 + T or 14.5 – \(\tau\) |
| (iv) | M1 | For using \(a = \frac{(0 - V)}{(14.5 - 8.5)}\) or \(0 = V + a(14.5 - 8.5)\) |
| Deceleration is 1 ms⁻² | A1ft | [2] |
**(i)** $[\frac{1}{2} \times 2.5(\text{speed}_{\max}) = 4]$ | M1 | For using area property for distance
Greatest speed is 3.2 ms⁻¹ | A1 | [2]
**SR(max. 1/2)** for candidates who (implicitly) make the unjustifiable assumption that $\text{speed}_{\max}$ occurs when $t = 1.25$
Greatest speed is 3.2 ms⁻¹ from $2 \times \frac{1}{2} \times 1.25(\text{speed}_{\max}) V = 4$ | B1 |
**(ii)** $[V = 3x2]$ | M1 | For using $a = \frac{(V - 0)}{(4.5 - 2.5)}$ or $V = 0 +$ at
$V = 6$ | A1 | [2]
**(iii)** | M1 | For using area property for distance from $t = 2.5$ to $t = 14.5$
$\frac{1}{2} \times 6(12 + T) = 48$ or $\frac{1}{2} \times 6x2 + 6T + \frac{1}{2} \times 6(10 - T) = 48$ or $\frac{1}{2} \times 6x2 + 6(10 - \tau) + \frac{1}{2} \times 6\tau = 48$ | A1ft |
$t = 8.5$ | A1 | [3] from 4.5 + T or 14.5 – $\tau$
**(iv)** | M1 | For using $a = \frac{(0 - V)}{(14.5 - 8.5)}$ or $0 = V + a(14.5 - 8.5)$
Deceleration is 1 ms⁻² | A1ft | [2]
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{881993e1-71ea-4801-bfc8-40c17a1387a9-3_579_1518_258_315}
The diagram shows the velocity-time graph for a particle $P$ which travels on a straight line $A B$, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the velocity of $P$ at time $t \mathrm {~s}$. The graph consists of five straight line segments. The particle starts from rest when $t = 0$ at a point $X$ on the line between $A$ and $B$ and moves towards $A$. The particle comes to rest at $A$ when $t = 2.5$.\\
(i) Given that the distance $X A$ is 4 m , find the greatest speed reached by $P$ during this stage of the motion.
In the second stage, $P$ starts from rest at $A$ when $t = 2.5$ and moves towards $B$. The distance $A B$ is 48 m . The particle takes 12 s to travel from $A$ to $B$ and comes to rest at $B$. For the first 2 s of this stage $P$ accelerates at $3 \mathrm {~m} \mathrm {~s} ^ { - 2 }$, reaching a velocity of $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find\\
(ii) the value of $V$,\\
(iii) the value of $t$ at which $P$ starts to decelerate during this stage,\\
(iv) the deceleration of $P$ immediately before it reaches $B$.\\
$7 \quad$ A particle $P$ travels in a straight line. It passes through the point $O$ of the line with velocity $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at time $t = 0$, where $t$ is in seconds. $P$ 's velocity after leaving $O$ is given by
$$\left( 0.002 t ^ { 3 } - 0.12 t ^ { 2 } + 1.8 t + 5 \right) \mathrm { m } \mathrm {~s} ^ { - 1 }$$
The velocity of $P$ is increasing when $0 < t < T _ { 1 }$ and when $t > T _ { 2 }$, and the velocity of $P$ is decreasing when $T _ { 1 } < t < T _ { 2 }$.\\
(i) Find the values of $T _ { 1 }$ and $T _ { 2 }$ and the distance $O P$ when $t = T _ { 2 }$.\\
(ii) Find the velocity of $P$ when $t = T _ { 2 }$ and sketch the velocity-time graph for the motion of $P$.
\hfill \mbox{\textit{CAIE M1 2010 Q6 [9]}}