Physics and math nascar question take a look?

Question

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straight-away at 74.0 m/s. The driver of the Thunderbird realizes that she must make a pit stop, and she smoothly slows to a stop over a distance of 250 m. She spends 5.00 s in the pit and then accelerates out, reaching her previous speed of 74.0 m/s after a distance of 410 m. At this point how far has the Thunderbird fallen behind the Mercedes Benz, which has continued at a constant speed?

Answer 1

The Thunderbird:

from the start of the race till 250m:

v^2=v0^2+2a(s-s0)

0=(74^2)+(2*a*250)

--> a= -10.95 m/s^2

v=v0+at

0=74-(10.95*t)

--> t1=6.76s

t2=5s (the time she spent stoped)

from 250m to 660m: (410m distance)

v^2=v0^2+2a(s-s0)

74^2=0+(2*a*410)

--> a=6.68m/s^2

v=v0+at

74=0+(6.68*t)

--> t3=11.08s

t(total)= 6.76+5+11.08= 22.84s

for the Benz:

s=s0+v0t+(0.5*a*(t^2))

s=0+74*(22.84)+0

=1690.16m

The distance between the two cars equals to:

1690.16-660=1030.16 m the thunderbird is behind.

I've used Newton's laws for constant accelaration.