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1. iam really stuck with this , iam not sure what is he asking for but i have some thoughs but none of them seem to work out.

two Cyclists c and d are travelling with constant
velocities (5i-2j)ms^-1 and 8jms^-1 respectively relative to a fixed origin O.

I solved (a) and (b) but iam gonna put them just for refrence.
(a) find the velocity of C relative to d.
at noon, the position vectors of C and d are (100i + 300j)m and (150i +100j) m respecively , referred to O. at t seconds after noon, the position vector of C relative to is s meters.
(b) show that s = (-50 + 5t)i + (200 - 10t)j.
this is the one.
(c) By Considering |s|^2 , or otherwise, find the value of t for which c and d are closest togeher.
2. So I guess you know how to find the modulus of a vector, yeah? |s|2 = |x|2 + |y|2.

So in this case |s|2 = (-50+5t)2 + (200-10t)2.

Now all you have to do is do d(|s|2)/dt = 0 and solve for t.
4. (Original post by Codefusion)
iam really stuck with this , iam not sure what is he asking for but i have some thoughs but none of them seem to work out.

two Cyclists c and d are travelling with constant
velocities (5i-2j)ms^-1 and 8jms^-1 respectively relative to a fixed origin O.

I solved (a) and (b) but iam gonna put them just for refrence.
(a) find the velocity of C relative to d.
at noon, the position vectors of C and d are (100i + 300j)m and (150i +100j) m respecively , referred to O. at t seconds after noon, the position vector of C relative to is s meters.
(b) show that s = (-50 + 5t)i + (200 - 10t)j.
this is the one.
(c) By Considering |s|^2 , or otherwise, find the value of t for which c and d are closest togeher.
for b)

s= u x t ... since they are travelling with constant velocities... then u subtract d from c ... and u will get ur answer
5. thank you very much. got it now kk

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