Question

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?

Answer 1

Thunderbird is 995.157 meters behind the Mercedes

Step-by-step 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
`v=0 \ m/s`

initial velocity
`u= 71 m/s` distance

Distance covered in the slowing down phase =
`250 m`

`v^2-u^2=2as`

`a= \frac {(v^2-u^2)}{2s}`

`a = \frac {(0^2-71^2)}{(2 * 250)}=-10.082 \ m/s^2`

`v=u+at`

`t= \frac {(v-u)}{a}`

`= \frac {(0-71)}{(-10.082)}=7.042 s`

`t_1=7.042 s`

The car is in the pit stop for 5s
`t_2=5 s`

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

`a= \frac {(v^2-u^2)}{(2* s)} = ((71^2-0^2))/((2 * 370)) =6.81 \ m/s^2`

`v=u+at`

`t= ((v-u))/(a)`

`t= ((71-0))/(6.81)= 10.425 s`

`t_3=10.425 s`

total time=
`t_1 +t_2+t_3=7.042+5+10.425=22.467 s`

Distance covered by the Mercedes Benz during this time is given by
`s=vt=71 * 22.467= 1595.157 m`

Distance covered by the Thunderbird during this time=
`250+350=600 m`

Difference between distance covered by the Mercedes and Thunderbird

=
`1595.157-600=995.157 m`

Thus the Mercedes is 995.157 m ahead of the Thunderbird.