In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straight-away at 69.5 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 69.5 m/s after a distance of 380 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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March 21st, 2011 at 1:26 pm
Taking the point at which the Thunderbird starts to stop as the origin, the distance the Mercedes travels is 69.5*t m
The distance the Thunderbird travels is 250 m to go from 69.5 m/s to 0; the average velocity over that period is 69.5/2, and the time is 250*2/69.5 = 7.19 s Then add 5 s in the pit to get 12.19 s. The time to accelerate to 69.5 m/s is 380*2/69.5 = 10.94 s giving a total time of 23.12 s. The distance traveled by the Thunderbird is 250+380 = 630 m
The distance traveled by the Mercedes is 23.12*69.5 = 1607 m. So the Thunderbird is 1607 - 630 = 977 m
Note: you can also find the stopping and startup time from s =0.5*a*t², the formula for distance traveled under uniform acceleration and initial velocity = 0. v = a*t, so a = v/t; plugging in for a gives s = 0.5*v*t and solving for t gives t = 2*s/v, the same as distance divided by average velocity