I stink at science, can u help me with these problems i have on webassign. #3?

Question

A speeder passes a parked police car at 35.0 m/s. The police car starts from rest with a uniform acceleration of 2.88 m/s2.
(a) How much time passes before the speeder is overtaken by the police car?
_____s
(b) How far does the speeder get before being overtaken by the police car?
_____m
-Next question-
A rocket moves upward, starting from rest with an acceleration of 30.2 m/s2 for 8.00 s. It runs out of fuel at the end of the 8.00 s but does not stop. How high does it rise above the ground?
_____m
-Next question-
A ball is thrown vertically upward (assumed to be the positive direction) with a speed of 22.0 m/s from a height of 3.0 m.
(a) How high does the ball rise from its original position?
_____m
(b) How long does it take to reach its highest point?
_____s
(c) How long does the ball take to hit the ground after it reaches its highest point?
_____s
(d) What is the ball's velocity when it returns to the level from which it started?
_____m/s
Thanks

Hey, um, thx for the answer, i hope u still read this, lol, but can you specify the answer, then i can work back and see how you got it.

Answer 1

Assuming the speeder maintains a constant velocity, the time of overtake is t, and the distance is d

d=35*t
and
d=.5*2.88*t^2

35*t=1.44*t^2
(note that at t=0 the equation is satisfied, this is because that is when the speeder passes the resting police car)
t=35/1.44
=24.3 seconds
d=850 m

for the ball
y(t)=3+22.0*t-.5*9.81*t^2
v(t)=22.0-9.81*t
determine t at apogee
v(t)=0
t=22/9.81
t=2.24 seconds (this is b)
to calculate a:
y(2.24)=3+22*2.24-.5*9.81*2.24^2
=27.7 m

for c, when is y(t)=0?
there will be two roots, the negative will be as if the ball was shot from ground level at a higher velocity) we want the positive root
t=4.62. This is the time from take-off. The questions asks
4.62-2.24
=2.38 seconds after apogee

d) is -22.0 m/s. This is by inspection because of the symmetry of the motion. It can also be calculate by finding the time when y(t)=3. t=0 is one answer, this is the start, and 4.49 seconds
v(4.49)=22-9.81*4.49
=-22.0 m/s

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