Simple Physics Help (SUVAT)

I have an exam tomorrow and i cant seem to work out there questions:
I know the answer but i dont know how to get to them!

6. A particle moves with uniform acceleration 0.5m/s2 in a horizontal line ABC. The speed of the particle at C is 80m/s and the times taken from A to B and from B to C are 40 and 30 seconds respectively. Calculate
(a) Speed at A
(b) Distance BC

Answer = (a) 45 ms-1
(b) 2175m

10. A train travels along a straight piece of track between 2 stations A and B. The train starts from rest at A and accelerates at 1.25m/s2 until it reaches a speed of 20m/s. It then travels at this speed for a distance of 1560m and then decelerates at 2m/s2 to come to rest at B. Find
(a) Distance from A to B
(b) Total time taken for the journey
(c) Average speed for the journey

Answer: (a) 1820m
(b) 104 secs
(c) 17.5ms-1

OK in a UVSAT question you want to start nailing down the unknowns.

here you know a = 0.5. v; the final velocity (occuring at C) is 80ms-1. The initial velocity u is the velocity at A. Another variable you know is t which is 40 seconds from A to B and 30 seconds from B to C. i.e. the time from A to C is 70 seconds.....

You can now plug that into the formulae: v = u + at

80 = u + 0.5 x 70. Therefore u = 80 - (0.5 x 70) = 80 - 35 = 45ms-1




The distance BC:

First find the velocity at B. v = u + at. Here you know that u = 45ms-1. you also know a = 0.5 and t = 40.

So v = u + at = 45 + 0.5 x 40 = 45 + 20 = 65ms-1

Now you know you start at 65ms-1 accelerate at 0.5ms-2 for 30 seconds and end up at 80ms-1. You need to know how far you go. [Check 80 - 30x0.5 = 65]

To do that you use the equation s = ut + 1/2at^2 (note you don't need to know v = 80 for this)

s is what you are trying to find.
u = 65
t = 30
a = 0.5

So s = 65 x 30 + 0.5 x 0.5 30^2 = 1950 + 225 = 2175