Physics help again please!!?

Question

1. In a 100 meter race Maggie and Judy cross the finish line in a dead heat, both taking 10.2 seconds. They both accelerated uniformly and Maggie took 2s and Judy took 3 to attain max. speed, which they maintained for the rest of the race. a)what was the acceleration of each. b)what were their max. speeds? c)which sprinter was ahead at the 6.0 second mark and by how much? 2. a jewel thief steals a golden egg from the third floor of an office building. Police enter and the thief runs to hide on an upper floor. To be free of the evidence, the thief drops the egg out the window. A detective on the 3rd floor sees the egg falling and times it as it passed the lower half of her window. The floors of the building are located 4 m apart. The egg took 0.101 s to fall 1.50 m past the lower part of the detective's window. The bottom of the detective's window is 8 m above the ground. On what floor is the thief hiding??
Anyone that can help I would really appreciate it. Thanks in advance!

Please help me. 10 points to pretty much anyone who gives me legitimate advice on how to solve this problem!

PLEASE HELP

Answer 1

1. For Maggie
the first 2 s are
vm(t)=am*t
and xm(t)=.5*am*t^2

for 2 vm(t)=am*2
and
xm(t)=.5*am*4+am*2*(t-2)
we know that when xm=100, t=10.2
solve for am
100=2*am*(t-1)

am=50/9.2 m/s^2
am=5.43 m/s^2

and the max speed is am*2
or 10.86 m/s

solve for Judy the same way

For Judy
the first 3 s are
vj(t)=aj*t
and xj(t)=.5*aj*(t^2)


for 3 vj(t)=aj*3
and
xj(t)=.5*aj*9+aj*3*(t-3)
we know that when xj=100, t=10.2
solve for aj
100=aj*(4.5+7.2*3)
aj=3.83


and the max speed is aj*3
or 11.5 m/s

now look at t=6
for Maggie
x(6)=.5*5.43*4+5.43*2*4

54.3 m

For Judy
xj(6)=.5*3.83*9+3.83*3*3
51.7 m

Maggie is ahead at t=6

2) The fall to the top of the window is
y(top)=y0+8-.5*g*t1^2
the speed of the egg is
v(top)=g*t1

the fall to the bottom of the window
y(bottom)=y(top)-v(top)*0.101
-.5*g*(0.101)^2

we also know that
y(top)-y(bottom)=1.50
and
y(top)-y(bottom)=
v(top)*0.101+.5*g*(0.101)^2

or
1.5=g*t1*0.101+
.5*g*(0.101)^2
solve for t1

and then the distance of the drop above the ground is
y0


j