M1 January 2008 Q6
6. [In this question, the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are due east and due north respectively.]
A particle \(P\) is moving with constant velocity \((-5\mathbf{i} + 8\mathbf{j})\) m s\(^{-1}\). Find
(a) the speed of \(P\), (2)
(b) the direction of motion of \(P\), giving your answer as a bearing. (3)
At time \(t = 0\), \(P\) is at the point \(A\) with position vector \((7\mathbf{i} - 10\mathbf{j})\) m relative to a fixed origin \(O\). When \(t = 3\) s, the velocity of \(P\) changes and it moves with velocity \((u\mathbf{i} + v\mathbf{j})\) m s\(^{-1}\), where \(u\) and \(v\) are constants. After a further 4 s, it passes through \(O\) and continues to move with velocity \((u\mathbf{i} + v\mathbf{j})\) m s\(^{-1}\).
(c) Find the values of \(u\) and \(v\). (5)
(d) Find the total time taken for \(P\) to move from \(A\) to a position which is due south of \(A\). (3)
| Scheme | Marks |
|---|---|
| Speed \(= \sqrt{(5^2 + 8^2)} \approx 9.43\) m s\(^{-1}\) | M1 A1 |
| (2) |
| Scheme | Marks |
|---|---|
| Forming arctan 8/5 or arctan 5/8 oe | M1 |
| Bearing \(= 360 - \arctan 5/8\) or \(270 + \arctan 8/5 = 328\) | DM1 A1 |
| (3) |
| Scheme | Marks |
|---|---|
| At \(t = 3\), p.v. of \(P = (7 - 15)\mathbf{i} + (-10 + 24)\mathbf{j} = -8\mathbf{i} + 14\mathbf{j}\) | M1 A1 |
| Hence \(-8\mathbf{i} + 14\mathbf{j} + 4(u\mathbf{i} + v\mathbf{j}) = \mathbf{0}\) | M1 |
| \(\Rightarrow\ \ u = 2,\ \ v = -3.5\) | DM1 A1 |
| (5) |
| Scheme | Marks |
|---|---|
| p.v. of \(P\) \(t\) secs after changing course \(= (-8\mathbf{i} + 14\mathbf{j}) + t(2\mathbf{i} - 3.5\mathbf{j})\) | M1 |
| \(= 7\mathbf{i} + \ldots\) | DM1 |
| Hence total time \(= 10.5\) s | A1 |
| (3) | |
| (13 marks) |