| Answer | Marks |
|---|---|
| 6 | (a) Direction of v = (7i – 7.5j) – (4i – 6j) = 3i – 1.5j M1 |
Question 6:
6 | (a) Direction of v = (7i – 7.5j) – (4i – 6j) = 3i – 1.5j M1
↓
1.5
tan θ = = 0.5 ⇒ θ = 26.565… M1 A1
3
Bearing = 117 (accept awrt) A1
(4)
(b) v = (3i – 1.5j) ÷ 3 = 4i – 2j B1
4
s = (4i – 6j) + t(4i – 2j) M1 A1√
(3)
(c) At 1015 s = (4i – 6j) + 5 (4i – 2j) ( = 9i – 8.5j) M1 A1
4
m = 0.25 (pi + qj) B1
↓
s = m ⇒ p = 36, q = – 34 M1 A1, A1
(6)
(a) Forming direction for v can be either way round.
M1 for tan = ‘i/j’ or ‘j/i’
A1 for 26.6 or 63.4 (awrt) from a correct direction for v
A1 cao
(b) Allow B1 for correct vector for v wherever seen (e.g. in (a))
(c) line 1: or (7i – 7.5j) + ½(4i – 2j) = …..
1st M1 allow for a valid attempt with a value of t.
2nd M1 using s = m and equating at least one coefficientA small boat $S$, drifting in the sea, is modelled as a particle moving in a straight line at constant speed. When first sighted at 0900, $S$ is at a point with position vector $(4\mathbf{i} - 6\mathbf{j})$ km relative to a fixed origin $O$, where $\mathbf{i}$ and $\mathbf{j}$ are unit vectors due east and due north respectively. At 0945, $S$ is at the point with position vector $(7\mathbf{i} - 7.5\mathbf{j})$ km. At time $t$ hours after 0900, $S$ is at the point with position vector $\mathbf{s}$ km.
\begin{enumerate}[label=(\alph*)]
\item Calculate the bearing on which $S$ is drifting. [4]
\item Find an expression for $\mathbf{s}$ in terms of $t$. [3]
\end{enumerate}
At 1000 a motor boat $M$ leaves $O$ and travels with constant velocity $(p\mathbf{i} + q\mathbf{j})$ km h$^{-1}$. Given that $M$ intercepts $S$ at 1015,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item calculate the value of $p$ and the value of $q$. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2004 Q6 [13]}}