2013 HCI P1 Q3

2013 HCI P1 Q3

6 marks
Prelims

There are two particles \(A\) and \(B\) with particle \(A\) at \(\left( -13,\ 0 \right)\) and particle \(B\) at \(\left( 0,\ -9 \right)\) with respect to origin \(O\). \(A\) moves towards \(O\) at a constant speed of \(u\ \mathrm{cm}\,{{\mathrm{s}}^{-1}}\) in the positive \(x\) direction and \(B\) moves towards \(O\) at a constant speed of \(\frac{4}{3}u\ \mathrm{cm}\,{{\mathrm{s}}^{-1}}\) in the positive \(y\)-direction (see diagram).

Show that, after a time of \(t\) seconds, the distance between \(A\) and \(B\) is given by

\(\sqrt{\left( \frac{25}{9}{{u}^{2}}{{t}^{2}}-50ut+250 \right)}\) cm.[2]

Find the coordinates of \(A\) and \(B\) at the instant when the two particles are closest.[4]

Video Solution:

Video Solution

Video solution locked

A free account saves your progress; this video needs tuition enrolment.

Solution:

Solution locked

680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs tuition enrolment.

Create a free account to practise similar questions.

Answer:\(AB=\sqrt{\frac{25}{9}u^2t^2-50ut+250}\text{ cm}\), with closest positions \(A=(-4,0)\) and \(B=(0,3)\)

Need help? Join our JC Math tuition classes.

Learn more