# Answer to Question #14623 in Mechanics | Relativity for CJ

Question #14623

n a quarter-mile drag race, two cars start simultaneously from rest, and each accelerates at a constant rate until it either reaches its maximum speed or crosses the finish line. Car A has an acceleration of 11.2 m/s2 and a maximum speed of 105 m/s. Car B has an acceleration of 11.9 m/s2 and a maximum speed of 85.9 m/s. (a) Which car wins the race? Type 1 for A, 2 for B. (b) By how many seconds does this car win the race?

Expert's answer

Car A

402.34m...................Space to cover

402.34 = 11.2*t^2/2....Space formula

t = √(2*402.34/11,2) = 8,476sec...time to goal

11.2*8,476 = 94,93m/sec.............top speed well below speed limit (105m/sec)

Car A can stretch the whole race at constant acceleration

Car B

t = √(2*402.34/11.9) = 8,223sec...time to goal

11.9*8,223 = 97,85m/sec.............top speed over speed limit (85.9 m/sec)

Car B can't stretch the whole race at constant acceleration; calculation approach must be different

85.9 m/sec......................Speed limit

85.9 /11.9 = 7,218sec....t1(time spent at constant acceleration)

1/2*11.9*7,218^2 = 309,99m........Space at const. acc

402.34-309,99= 92,35m...............Space at const. speed

92,35/85.9& = 1,075.....t2 (time spent at constant speed)

7,218+1,075= 8,293sec........total time to target

Car B wins the competition with an advantage worth:

-(8,293-8,476) = 0,183sec

402.34m...................Space to cover

402.34 = 11.2*t^2/2....Space formula

t = √(2*402.34/11,2) = 8,476sec...time to goal

11.2*8,476 = 94,93m/sec.............top speed well below speed limit (105m/sec)

Car A can stretch the whole race at constant acceleration

Car B

t = √(2*402.34/11.9) = 8,223sec...time to goal

11.9*8,223 = 97,85m/sec.............top speed over speed limit (85.9 m/sec)

Car B can't stretch the whole race at constant acceleration; calculation approach must be different

85.9 m/sec......................Speed limit

85.9 /11.9 = 7,218sec....t1(time spent at constant acceleration)

1/2*11.9*7,218^2 = 309,99m........Space at const. acc

402.34-309,99= 92,35m...............Space at const. speed

92,35/85.9& = 1,075.....t2 (time spent at constant speed)

7,218+1,075= 8,293sec........total time to target

Car B wins the competition with an advantage worth:

-(8,293-8,476) = 0,183sec

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