| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Distance between two moving objects |
| Difficulty | Standard +0.3 This is a standard M1 mechanics question on relative motion and kinematics. Part (a) requires simple position vector formulation, part (b) involves finding distance using Pythagoras, and part (c) requires minimization by completing the square or differentiation. All steps are routine applications of well-practiced techniques with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 1.10f Distance between points: using position vectors1.10h Vectors in kinematics: uniform acceleration in vector form3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks |
|---|---|
| (a) \(\vec{r}_P = (\vec{i} + 7\vec{j}) + t(3\vec{i} - 4\vec{j}) = (1 + 3t)\vec{i} + (7 - 4t)\vec{j}\) | M1 A1 A1 |
| \(\vec{r}_Q = (3\vec{i} - 8\vec{j}) + t(6\vec{i} + 5\vec{j}) = (3 + 6t)\vec{i} + (5t - 8)\vec{j}\) | M1 A1 A1 |
| (b) \(\vec{PQ} = (2 + 3t)\vec{i} + (9t - 15)\vec{j}\) | M1 A1 A1 |
| \(PQ = \sqrt{(2 + 3t)^2 + (9t - 15)^2}\) | M1 A1 |
| \(= \sqrt{90t^2 - 258t + 229}\) | |
| (c) \(\frac{d}{dt}(PQ^2) = 180t - 258 = 0\) for min. | M1 A1 M1 A1 |
| \(t = 1.43 \text{ hrs} = 86 \text{ mins}\) | |
| so time is \(1:26 \text{ p.m.}\) | A1 A1 |
| Then \(PQ = \sqrt{44.1} = 6.64 \text{ km}\) | |
| 17 marks |
**(a)** $\vec{r}_P = (\vec{i} + 7\vec{j}) + t(3\vec{i} - 4\vec{j}) = (1 + 3t)\vec{i} + (7 - 4t)\vec{j}$ | M1 A1 A1 | |
$\vec{r}_Q = (3\vec{i} - 8\vec{j}) + t(6\vec{i} + 5\vec{j}) = (3 + 6t)\vec{i} + (5t - 8)\vec{j}$ | M1 A1 A1 | |
**(b)** $\vec{PQ} = (2 + 3t)\vec{i} + (9t - 15)\vec{j}$ | M1 A1 A1 | |
$PQ = \sqrt{(2 + 3t)^2 + (9t - 15)^2}$ | M1 A1 | |
$= \sqrt{90t^2 - 258t + 229}$ | | |
**(c)** $\frac{d}{dt}(PQ^2) = 180t - 258 = 0$ for min. | M1 A1 M1 A1 | |
$t = 1.43 \text{ hrs} = 86 \text{ mins}$ | | |
so time is $1:26 \text{ p.m.}$ | A1 A1 | |
Then $PQ = \sqrt{44.1} = 6.64 \text{ km}$ | | |
| | | 17 marks |\begin{enumerate}
\item At noon, two boats $P$ and $Q$ have position vectors $( \mathbf { i } + 7 \mathbf { j } ) \mathrm { km }$ and $( 3 \mathbf { i } - 8 \mathbf { j } ) \mathrm { km }$ respectively relative to an origin $O$, where $\mathbf { i }$ and $\mathbf { j }$ are unit vectors in the directions due East and due North respectively. $P$ is moving with constant velocity $( 3 \mathbf { i } - 4 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }$ and $Q$ is moving with constant velocity $( 6 \mathbf { i } + 5 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }$.\\
(a) Find the position vector of each boat at time $t$ hours after noon, giving your answers in the form $\mathrm { f } ( t ) \mathrm { i } + \mathrm { g } ( t ) \mathrm { j }$, where $\mathrm { f } ( t )$ and $\mathrm { g } ( t )$ are linear functions of $t$ to be found.\\
(b) Find, in terms of $t$, the distance between the boats $t$ hours after noon.\\
(c) Calculate the time when the boats are closest together and find the distance between them at this time.
\item A particle starts from rest and accelerates at a uniform rate over a distance of 12 m . It then travels at a constant speed of $u \mathrm {~ms} ^ { - 1 }$ for a further 30 seconds. Finally it decelerates uniformly to rest at $1.6 \mathrm {~ms} ^ { - 2 }$.\\
(a) Sketch the velocity-time graph for this motion.\\
(b) Show that the total time for which the particle is in motion is
\end{enumerate}
$$\frac { 5 u } { 8 } + 30 + \frac { 24 } { u } \text { seconds. }$$
(c) Find, in terms of $u$, the total distance travelled by the particle during the motion.\\
(d) Given that the total time for the motion is $39 \cdot 5$ seconds, show that $5 u ^ { 2 } - 76 u + 192 = 0$.\\
(e) Find the two possible values of $u$ and the total distance travelled in each case.
\hfill \mbox{\textit{Edexcel M1 Q6 [17]}}