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# M2 watch

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1. if u use t=4
then v=8i+4j
which is in the same ratio thus parallel to 2i+j
2. As youve probably guessed Im struggling with this
so:

A particle X, moving along a straight line with constant speed 4m/s, passes through a fixed point O. Two seconds later another particle Y, moving along the same straight line and in the same direction, passes through O with speed 6m/s.
Given that Y is subject to a cosntant deceleration of magnitude 2m/s/s

a) state the velocity and displacement of each particle t seconds after Y passed through O

b) find the shorest distance between the particles after they have both passed through O

c) find the value of t when the distance between the particle has increased to 23m

I struggled through a, didnt really understand it though, and dont know what to do for b/c. Can anyone go through it for me please
3. Bump
4. (Original post by imasillynarb)
As youve probably guessed Im struggling with this
so:

A particle X, moving along a straight line with constant speed 4m/s, passes through a fixed point O. Two seconds later another particle Y, moving along the same straight line and in the same direction, passes through O with speed 6m/s.
Given that Y is subject to a cosntant deceleration of magnitude 2m/s/s

a) state the velocity and displacement of each particle t seconds after Y passed through O
i'm not too sure if i've done this right, but here's a) :

vX = 4m/s
vY = (6-2t) m/s

when t=2, rX = 8m , rY = 0m
so at time t seconds, rX = (8 + 4t) m
rY = (6-2t)t = (6t - 2t^2)m
5. (Original post by imasillynarb)

b) find the shorest distance between the particles after they have both passed through O
i think you would just do (rX-rY)

D = 8+4t - (6t - 2t^2)
D = 2t^2 - 2t + 8

to find min. differentiate: dD/dt = 4t - 2 = 0

Therefore t=0.5 (to prove it's a min, differentitate again to get 4 => min)

sub t=0.5, D= 7.5m
6. for last one, make D = 23, and get a quadratic to solve t. i get t= [1+rt(31)]/2

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