(a) Two ships M and N, moving with constant velocities, have position vectors (3i + 7j) and (4i + 5j) respectively. If the velocities of M and N are (5i + 6...
Question 1 Report
(a) Two ships M and N, moving with constant velocities, have position vectors (3i + 7j) and (4i + 5j) respectively. If the velocities of M and N are (5i + 6j) and (2i + 3j) and the distance covered by the ships after t seconds are in metres, find (i) MN ; (ii) |MN|, when t = 3 seconds.
(b) A particle is acted upon by forces \(F_{1} = 5i + pj ; F_{2} = qi + j ; F_{3} = -2pi + 3j\) and \(F_{4} = -4i + qj\), where p and q are constants. If the particle remains in equilibrium under the action of these forces, find the values of p and q.
Answer Details
(a) After time \(t\), position \(=\) initial position \(+\) velocity\(\times t\):
\[M: (3+5t)\mathbf{i} + (7+6t)\mathbf{j},\qquad N: (4+2t)\mathbf{i} + (5+3t)\mathbf{j}.\]
(i) \(\overrightarrow{MN} = \) position of \(N\) minus position of \(M\):
\[\overrightarrow{MN} = (4+2t-3-5t)\mathbf{i} + (5+3t-7-6t)\mathbf{j} = (1 - 3t)\mathbf{i} - (2 + 3t)\mathbf{j}.\]
(ii) At \(t = 3\): \(\overrightarrow{MN} = (1-9)\mathbf{i} - (2+9)\mathbf{j} = -8\mathbf{i} - 11\mathbf{j}\), so
\[|\overrightarrow{MN}| = \sqrt{(-8)^2 + (-11)^2} = \sqrt{64 + 121} = \sqrt{185} \approx 13.6\ \text{m}.\]
(b) For equilibrium the sum of the forces is zero. Adding components:
\(\mathbf{i}\): \(5 + q - 2p - 4 = 0 \Rightarrow q = 2p - 1\).
\(\mathbf{j}\): \(p + 1 + 3 + q = 0 \Rightarrow q = -p - 4\).
Equating: \(2p - 1 = -p - 4 \Rightarrow 3p = -3 \Rightarrow p = -1\), and \(q = -(-1) - 4 = -3\).
Answer: \(p = -1,\; q = -3\).
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