In the Daytona 500 auto race, a Ford Thunderbird and a MercedesBenz are moving side by side down a straightaway at71.5 m/s. The driver of the Thunderbird realizes that she mustmake a pit stop, and she smoothly slows to a stop over a distanceof 250 m. She spends 5.00 s in the pit and then accelerates out,reaching her previous speed of 71.5 m/s after a distance of 350m. At this point, how far has the Thunderbird fallen behind theMercedes Benz, which has continued at a constant speed?

Question
Asked Dec 16, 2019
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In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes
Benz are moving side by side down a straightaway at
71.5 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 71.5 m/s after a distance of 350
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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Expert Answer

Step 1

Given:

Velocity of Ford Thunderbird = 71.5 m/s

Velocity of Mercedes Benz = 71.5 m/s

Step 2

Thunderbird stops over a distance of 250 m. After 5 s she again accelerates to regain its original velocity after covering a distance of 350 m. Deacceleration and acceleration for these can be calculated by using equations of motion.

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for 1st case: v} =v} +2a,s 0=(71.5) +2a, x 250 5112.25 500 a, = -10.22m/s %3D %3D 0-71.5 -10.22 t; = 7s for 2nd case: v} =v} +2a,s (71.5) = 0+ 2a, x350 5112.25 = a, 700 a, = 7.30m/s 1, = a. 71.5-0 7.30 t = 9.8s

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Step 3

Time taken by her to...

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t =t, +t, t =7+5+9.8 t=t, +t, rest t = 21.8s

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