Answer to Question #47244 in MatLAB | Mathematica | MathCAD | Maple for Sohil
The car A accelerates during time t_a1 = v1/a1 = 109/11.7 = 9.31 s and covers a distance s1 = a1*t_a1^2/2 = 507 m. Thus, the total time of motion is T1 = sqrt(2L/a1) = 8.28 s.
In the same way car B accelerates during time t_a2 = 92/11.1 = 8.29 s and covers a distance s2 = 381 m. t_a2 < T1, thus the car A wins.
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Thank you, we corrected the answer.
There was a track distance given, it says a quarter of a mile. That makes this answer not correct.