Mechanics-constant acceleration

"A bus pulls away from a stop with an acceleration of 1.5 m/s^2 which is maintained until the speed reaches 12 m/s. At the same instant a girl who is 5m away from the stop starts to run after the bus at a constant 7m/s. Will the girl catch the bus?"

This is how i have started it:

variables for bus ----> a = 1.5; v = 12
Variables for girl-----> s =5; v = 7

The bus starts from a stop, so initially v=0. The v=12 information is actually completely irrelevant since the girl can't possibly catch the bus when it's moving that fast whether it continues to accelerate or not.
The girl has acceleration a=0.

The equation should be s = vt + 1/2at^2 (a positive acceleration will increase the distance moved, not decrease it) and I would use this equation for both. You could try using two variables s_girl and s_bus. You want to find if there's a point in time when these two values are the same. Can you see what to do now? (Remember that the girl starts at -5m, what would you do to account for this?)

we can't use s = vt +1/2at^2 for the bus cause then we have two variables we need to solve right????

edit - got no idea what you mean by relative positions mate!!!!

By relative positions I mean s_bus-s_girl. If this is positive, the bus is still ahead of the girl. If it's zero (or negative) the girl has caught up. (Here I'm assuming the girl starts at -5 and they're both travelling in the positive direction.)

You don't necessarily have to solve for any variables unless there's a second part of the question. This is where s_bus-s_girl helps. At the moment she catches up, s_bus-s_girl=0. If there isn't any t that gives this to be 0 then she doesn't catch up. Using s = vt +1/2at^2 and considering s_bus-s_girl=0 should give you a quadratic.

Okay, so basically, we need s = vt+1/2at^2 for both but i don't get why we can't do it my way???? cause the answer is "yes and after 0.78s" so i know that if i get the time for the bus and then get the time for girl, if the girl's time is less than the bus therefore it has caught up hence why i did v = u +at for the bus and got the time as 8 seconds..... but i can't get "t" for the girl..... soz about that, maybe if we start there then your posts will make more sense to me.

The time for them do do what? Presumably the time for them to reach a particular point. But how do you decide upon the point?

You got 8 seconds for the bus to reach 12m/s and you could then work out how long it took the girl to reach wherever the bus has got to after 12 seconds. If she took longer you don't know whether she got ahead at some point and then fell behind later, so you still don't know what the answer is. By chance it works for this question, but only just (the bus gets ahead again before 9 seconds).

If you're going to check one point in this manner you should check how far the bus has got by the time it's reached 7m/s as this is when the girl stops catching up. This is the only value of v you can choose that'll guarantee you get the answer. Doing this you can sub 7m/s into v=u+at (for the bus), sub t into s = vt +1/2at^2 (for the bus) and then sub s into s = ut (for the girl) to find that the girl actually reaches this point first. This seems to me to be the more complicated approach though.

Okay, that is the way i wanted tbh cause it makes more sense to me. So i use v = u +at for bus to find t and then find s using s = ut +1/2at^2 and now once i got that....... oh i see, i get it!!!!!

edit - i dont' get the answer of 0.78s for the girl though using that method??????

You can't work out t using that method because you need to pick a speed, and there's no guarantee you'll pick the right one unless you already know the answer. That's not quite true actually: you could work out t, but you'd have to try every single possible distance until you get two distances where their times are the same. If you use that method it would be the shorter of the two distances, but I strongly advise against it. The question doesn't ask for t though, so you can use the pick a speed/distance method if you want, you just can't expect to get the value of t at which she catches up with the bus.

Do you mind showing me how you got the answer then??? from there i can see the errors of my way.

Which method do you want me to work through? The one I originally suggested?

Yes please, the one you originally suggested!!!!

Here you go:

s=ut+1/2at²
s_bus=0.75t²
s_girl=7t-5 (since she starts 5m behind)
s_bus-s_girl=0.75t²-7t+5
We want to find t such that this difference is zero.
0.75t²-7t+5=0
t²-(28/3)t+20/3=0 (multiplying through by 4/3 to make it easier)
(t-14/3)²=136/9 (completing the square, you could obviously use the quadratic formula instead)
Note that if the right hand side was negative there wouldn't be a solution so she would never catch up.
t-14/3=(+/-)(136/9)^1/2
We want the smaller answer (the bigger answer is when the bus would overtake again if they kept on going), so choose the minus sign.
t=14/3-(136/9)^1/2
t=0.78 seconds

i actually got 0.75t^2 -7t -5 which is wrong but along the same lines with a totally different method (the one i originally did first) but this is perfect and exactly what i need. Thanks mate for your help, it is much appreciated, +rep for you thanks once again!!!

"A particle P, moving along a straight line with constant acceleration 0.3 ms^-2 passes a point A on the line with a velocity of 20 ms^-1. At the instant when P passes A, a second particle Q is 20m behind A and moving with velocity 30ms^-1. Prove that, unless the motion of P and/or Q changes, the particles will collide."

I did, s = vt + 1/2at^2

Sp = 20t + 0.15t^2
Sq = 30t - 20

If they do collide one of their times will have to equal one another..... so just solve the equation really

Sq----> 30t - 20 = 0.....t = 2/3
Sp-----> 0.15t^2+20t = 0.... t(0.15t -20)...... t = 0 and t = 133s

So i am totally off here, where am i going wrong?????

edit - do you do this -----> 0.15t^2 + 20t = 30t - 20

0.15t^2 - 10t + 20 = 0..... solve for t and then sub that in for both equations and the S will be the same right?