Problem
Question 1 (5 marks) Two particles $\mathrm{P}$ and $\mathrm{Q}$ are moving along a horizontal line. At any time $t$ seconds, the position of particle $\mathrm{P}$ is given by $x=e^{t}$ and the position of the particle $\mathrm{Q}$ is given by $x=1+6 e^{-t}$. The diagram below shows the position of particle $\mathrm{P}, x$ metres, at any time $t$ seconds. (a) As time increases indefinitely, what position does the particle Q approach? (b) Sketch the path of particle $Q$ on a cartesian plane. (c) Calculate the position where the two particles meet. (d) Explain why $\mathrm{P}$ and $\mathrm{Q}$ will never travel at the same velocity.
Solution
Step 1 ：\(x_P = e^t\) and \(x_Q = 1 + 6e^{-t}\)
Step 2 ：(a) \(\lim_{t \to \infty} x_Q = \lim_{t \to \infty} (1 + 6e^{-t}) = 1\)
Step 3 ：(b) Sketch the path of particle Q on a cartesian plane.
Step 4 ：(c) \(e^t = 1 + 6e^{-t}\)
Step 5 ：\(e^{2t} = 6\)
Step 6 ：\(2t = \ln 6\)
Step 7 ：\(t = \frac{1}{2} \ln 6\)
Step 8 ：\(x = e^{\frac{1}{2} \ln 6} = \sqrt{6}\)
Step 9 ：\(\boxed{(t, x) = (\frac{1}{2} \ln 6, \sqrt{6})}\)
Step 10 ：(d) \(v_P = \frac{dx_P}{dt} = e^t\) and \(v_Q = \frac{dx_Q}{dt} = -6e^{-t}\)
Step 11 ：Since \(e^t > 0\) and \(-6e^{-t} < 0\), \(v_P\) and \(v_Q\) will never be equal.
