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?
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Accessories March 21st 2011
March 21st, 2011 at 7:28 am
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.