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Vector mechanics - how to find value of t where two cyclists are closest together

I was wondering what the method is to find the value of t where two moving objects are closest together.
For example if s (the position vector of C relative to D) is:
s= (-50 + 5t)i + (200-10t) j, how would you find the value of t where C and D are closest together; apparently you should square s but not sure where to go from there???
Reply 1
Avatar for JSG29
JSG29
11
Finding minimum distance between 2 objects:
Use Pythagoras' theorem to find s2
Differentiate s2 with relation to t, and make that equal to 0
Solve for t

If you need further explanation/reasoning, feel free to ask

JG
Original post by mayjb
I was wondering what the method is to find the value of t where two moving objects are closest together.
For example if s (the position vector of C relative to D) is:
s= (-50 + 5t)i + (200-10t) j, how would you find the value of t where C and D are closest together; apparently you should square s but not sure where to go from there???


Right, okay.

You do not 'square' the ss. Since this is relative motion, and C is moving relatively to D whereby ss is their relative distance apart, you can use Pythagoras' theorem to work out the magnitude of the distance. Just as the magnitude of the vector a=3i+4ja=3\text{i}+4\text{j} is ∣a∣=32+42=5\lvert a \lvert = \sqrt{3^2+4^2}=5. Applying it to this question would give you:

∣s∣=(−50+5t)2+(200−10t)2⇒s2=(−50+5t)2+(200−10t)2\lvert s \lvert=\sqrt{(-50+5t)^2+(200-10t)^2} \Rightarrow s^2=(-50+5t)^2+(200-10t)^2

Now we need to find the time when they are closest together. This means when the change in distance over the amount of time taken is 0, or a minimum rather, hence differentiation. It is easy to visualise 2 objects moving, with a line connected them two, so the line would begin to decrease in length as the objects get closer together. At the shortest distance the line stops decreasing, and begins to increase in length instead as the two objects pass by and keep going. A bit like a parabola when you go along it you reach a point where you stop and start going in the opposite direction. This means we need to differentiate our equation with respect to tt using the chain rule, or otherwise, and set dsdt=0\frac{ds}{dt}=0 and solve for tt
(edited 7 years ago)
Reply 3
Avatar for IYGB
IYGB
11
Completing the square under the square root also works
Reply 4
Avatar for Zacken
Zacken
22
Original post by RDKGames
Since this is relative motion, and C is moving relatively to D whereby ss is their relative distance apart, you can use Pythagoras' theorem to work out the magnitude of the distance. Just as the magnitude of the vector


Distance is a scalar, displacement is a vector.
Reply 5
Avatar for Krollo
Krollo
17
A nice method to have in your toolkit is to recognise that if the relative displacement is r, we want to minimise r^2.

Differentiating with respect to t using the chain rule, this gives

2r.v=0,

I.e. the relative velocity and displacement are perpendicular.

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