In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at

67.0 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 67.0 m/s after a

distance of 360 m. At this point how far has the Thunderbird fallen behind the Mercedes Benz, which has continued at a

constant speed?

In this first motion, the initial velocity is given as u = 67 m/s, while the final velocity which is the stop is v = 0 m/s and the distance is s = 250 m

Thus;

a = (0² - 67²)/(2 × 250)

a = -4489/500

a = -8.978 m/s²

Now, to get the time spent for this 250 m distance, we will use Newton's 2nd law of motion;

s = ut + ½at²

Plugging in the relevant values;

250 = 67t + ½(-8.978)t²

250 = 67t - 4.489t²

4.489t² - 67t + 250 = 0

Using quadratic formula, we will arrive at;

t = 7.463 s

Now, we are told that she accelerates out, reaching her previous speed of 67.0 m/s after a distance of 360 m. Thus, let's calculate the time it took to accelerate after a distance of 360 m.

Since she stayed from rest in this phase, then initial velocity (u) = 0 m/s.

Like in the first phase, we can find the acceleration from;

a = (v² - u²)/2s

a = (67² - 0²)/(2 × 360)

a = 4489/720

a = 6.235 m/s²

Thus, we can find the time for this 360 m phase by using; s = ut + ½at², we have;

360 = 0 + ½(6.235)t²

Multiply through by 2 to get;

6.235t² = 720

t² = 720/6.235

t² = 115.477145

t = √115.477145

t = 10.746 s

We are told that she spends 5 s in the pit. Thus;

Total time is;

t_total = 7.463 + 5 + 10.746

t_total = 23.209 s

Distance covered by the Mercedes Benz = 67 × 23.209 = 1555 m

Total distance covered by Thunderbird = 250 + 360 = 610 m

Thus, distance of Thunderbird behind Mercedes Benz is = 1555 - 610 = 945 m