In the 1960s there was a contest to find the car that could do the following two maneuvers (one right after the other) in the shortest total time: First, accelerate from rest to 100 mi/h, (45.0 m/s), and then brake to a complete stop. (Ignore the reaction time correction that occurs between the speeding-up and slowing-down phases and assume that all accelerations are constant.) For several years, the winner was the “James Bond car,” the Aston Martin. One year it won the contest when it took a total of only 15.0 seconds to perform these two tasks! Its braking acceleration (deceleration) was known to be an excel-lent 9.00 m/s2.
(a) Calculate the time it took during the braking phase.
(b) Calculate the distance it traveled during the braking phase.
(c) Calculate the car’s acceleration during the speeding-up phase.
(d) Calculate the distance it took to reach 100 mi/h.

  • CreatedAugust 29, 2015
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