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Just need a little help with *another* accelerated linear motion question, any help is appreciated:

(A) A particle starts from rest at a point p and accelerates at 2m/s^2 until it reaches a speed v m/s.

It travels at this speed for one minute before delerating at 1m/s^2 to rest at q. The total time for the journey is 2 minutes.

(i) calculate the distance pq

(ii)If a second particle starts from p at time t=0 and moves along pq with speed (2t + 50) m/s, find the time taken to reach q.

Part one ive gotten out - pq = 3600m

Part two, im not really too sure where to start.

(B) A particle P is projected vertically upwards with an initial velocity u and two seconds later a second particle Q is projected vertically upwards from the same point with initial velocity 1.5u. Calculate, in terms of u, how long W is in motion before it collides with p, and prove that u > 9.8.

Ok, so ive made a bit of a start on this one, whether or not its right is what im trying to find out, and what i go about doing next:

Eqn1 : S=ut+(1/2)at^2

Eqn2 : S=ut-(1/2)gt^2

S=ut-(1/2)gt^2

s=(1.5u)(t+2) - (1/2)g(t+2)^2

s=1.5ut+3u-(1/2)g(t^2+4t+4)

s=1.5ut+3u-(1/2)gt^2-2g-2gt

Collision occurs when s=s

ut-(1/2)gt^2 = 1.5ut+3u-(1/2)gt^2-2g-2gt

ut=1.5ut+3u-2gt-2g

2ut+3ut+6u-4gt-4g

-ut+4gt=6u-4g

t(4g-u)=6u-4g

t= (6u-4g)/(4g-u)

So thats all that ive got so far, any help very appreciated

Thanks again