In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straight-away at 68.5 m/s. The driver of the Thunderbird realizes he must make a pit stop, and he smoothly slows to a stop over a distance of 250 m. He spends 5.00 s in the pit and then accelerates out, reaching his previous speed of 68.5 m/s after a distance of 340 m. At this point, how far has the Thunderbird fallen behind the Mercedes Benz, which has continued at a constant speed? (respond in meters)
_________m
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March 21st, 2011 at 8:04 pm
We need to consider this question in three parts: The Thunderbird’s deceleration phase, the period when he is in the pit and then his acceleration phase. We know the accelerating and decelerating is smooth, in other words the values of the acceleration is constant.
Deceleration phase:
We need to find the time it takes to decelerate over a distance of 250 m. We can use one of the kinematic equations, we have the following information:
d = distance
u = initial speed = 68.5 m/s
v = final speed = 0 m/s
t = time = ?
d = t(u+v)/2. Since v = 0, this reduces to d = tu/2, and therefore t = 2d/u = 2*250/68.5 = 7.299 s. We know that over that time the thunderbird has travelled 250 m. Next, find how far the Benz travels in that time. We know its speed is constantlt 68.5 m, so distance is just speed times time
= (7.299 s)(68.5 m/s) = 500 m. So during the deceleration phase, the Thunderbird loses 250 m.
In the Pit Stop:
We know the Thunderbird is at rest at that phase, so the distance lost will be equal to the distance the Benz travels during that time
= (5 s)(68.5 m/s)
= 342.5 m.
Acceleration phase:
Approach this part of the journey in a similar manner to the way used in the deceleration phase:
d = distance = 340 m/s
u = initial speed = 0 m/s
v = final speed = 68.5 m/s
t = time = ?
d = t(u+v)/2. Since u=0, d = tv/2 so t = 2d/v = 2*340/68.5 = 9.927 s. We know th Thunderbird has travelled 340 m in that time, whereas the Benz has travelled
(9.927 s)(68.5 m/s) = 680 m, so the Thunderbird loses a further 340 m during that phase.
So the total distance the Thunderbird loses is 250 m + 342.5 m + 340 m
= 932.5 m.