Science:Math Exam Resources/Courses/MATH100/December 2014/Question 07 (a)

MATH100 December 2014

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Question 07 (a)
Particle $1$ moves on the $y$-axis starting at the point $(0,6)$ and travels towards origin with constant speed of $2$ units per second. Particle $2$ moves on the $x$-axis starting at the origin and travels in the positive $x$-direction with constant speed of $1$ unit per second. Find a formula for the distance between the particles as a function of time.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
First write equations for the coordinates of each of the particles as a function of time.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Drawn is the starting position and velocity of the particles. Let $t$ be time in seconds. The position of the first particle is given by $(0,y)$ where $y=6-2t$ The position of the second particle is $(x,0)$ with $x=t$ The squared-distance between the particles is therefore ${\begin{aligned}d^{2}&=(\mathrm {difference\ in\ } x)^{2}+(\mathrm {difference\ in\ } y)^{2}\\&=t^{2}+(6-2t)^{2}\\&=5t^{2}-24t+36\\d&={\sqrt {5t^{2}-24t+36}}\end{aligned}}$