Two particles are fired together along a smooth horizontal surface with velocities 4 m/s and 5 m/s. If they move at 60° to each other, find the distance between them in 2 seconds.

  • A \(2\sqrt{61}\)
  • B \(\sqrt{42}\)
  • C \(2\sqrt{21}\)
  • D \(2\sqrt{10}\)
waec 2010furthermathematics

The correct answer is C. \(2\sqrt{21}\)

Given lines \(OA\) and \(OB\) inclined at angle \(\theta\), the line \(AB\) is gotten using cosine rule.

\(|AB|^{2} = |OA|^{2} + |OB|^{2} - 2|OA||OB|\cos \theta\)

\(|AB|^{2} = 4^{2} + 5^{2} - 2(4)(5)\cos 60\)

= \(16 + 25 - 20\)

\(|AB|^{2} = 21 \implies |AB| = \sqrt{21}\)

\(\implies \text{The two particles are} \sqrt{21} m \text{apart in 1 sec}\)

In two seconds, the particles will be \(2\sqrt{21} m\) apart.

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