Problem:
In the daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.5 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 250m . She spends 5.00 s in th pit and then accelerates out , reaching her previous speed of 71.5 m/s after a distance of 350 m . At this point , how far has the Thunderbird fallen behind the Mercedes Benz , which has continued at a constant speed?
Could you explain me step by step?
What’s the answer??
Thank youu
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Accessories March 20th 2011
March 20th, 2011 at 2:02 pm
There are three steps to consider
(1) The Ford slows down from 71.5 m/s to 0 m/s over a distance of 250 m.
Using v^2 = u^2 - 2as, this is an deceleration of 10.225 m s-2.
Using v = u + at, this takes a time of 6.993 s.
In this 6.993 seconds the M Benz has travelled 6.993 * 71.5 = 500 m, so has gained 250 m on the Ford
(2) While in the pits, the M Benz gains another 5 * 71.5 m on the Ford. 357.5 m.
(3) The TB then accelerates to 71.5 m/s over 350 m.
Using v^2 = u^2 - 2as, this is an acceleration of 7.303 m s-2.
Using v = u + at, this takes a time of 9.790 s.
In this 9.790 s the M Benz has travelled 9.790 * 71.5 = 700 m, so has gained 350 m on the Ford.
Overall, the M Benz gains 957.5 m on the Ford.
Check my working carefully. Draw this out on a time-distance graph to check.