In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 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 71.0 m/s after a distance of 370 m. At this point how far has the Thunderbird fallen behind the Mercedes Benz, which has continued at a constant speed?

Respuesta :

Answer:

Thunderbird is 995.157 meters behind the Mercedes

Explanation:

It is given that all the cars were moving at a speed of 71 m/s when the driver of Thunderbird  decided to take a pit stop and slows down for 250 m. She spent 5 seconds  in the pit stop.

Here final velocity [tex]v=0 \ m/s[/tex]

initial velocity [tex]u= 71 m/s[/tex]  distance  

Distance covered in the slowing down phase = [tex]250 m[/tex]

[tex]v^2-u^2=2as[/tex]

[tex]a= \frac {(v^2-u^2)}{2s}[/tex]

[tex]a = \frac {(0^2-71^2)}{(2 \times 250)}=-10.082 \ m/s^2[/tex]

[tex]v=u+at[/tex]

[tex]t= \frac {(v-u)}{a}[/tex]

[tex]= \frac {(0-71)}{(-10.082)}=7.042 s[/tex]

[tex]t_1=7.042 s[/tex]

The car is in the pit stop for 5s [tex]t_2=5 s[/tex]

After restart it accelerates for 350 m to reach the earlier velocity 71 m/s

[tex]a= \frac {(v^2-u^2)}{(2\times s)} = \frac{(71^2-0^2)}{(2 \times 370)} =6.81 \ m/s^2[/tex]

[tex]v=u+at[/tex]

[tex]t= \frac{(v-u)}{a}[/tex]

[tex]t= \frac{(71-0)}{6.81}= 10.425 s[/tex]

[tex]t_3=10.425 s[/tex]

total time= [tex]t_1 +t_2+t_3=7.042+5+10.425=22.467 s[/tex]

Distance covered by the Mercedes Benz during this time is given by [tex]s=vt=71 \times 22.467= 1595.157 m[/tex]

Distance covered by the Thunderbird during this time=[tex]250+350=600 m[/tex]

Difference between distance covered by the Mercedes  and Thunderbird

= [tex]1595.157-600=995.157 m[/tex]

Thus the Mercedes is 995.157 m ahead of the Thunderbird.