A. Two cars B drive in the same direction on the straight road. Car a moves at a constant speed at the speed of VA = 20m / s, and car B moves at a constant deceleration straight line at the initial speed of vb0 = 24m / s, with the acceleration of a = 2m / S2. If car B is at △ s = 165m in front of car a, how long does it take for the two cars to meet? A student has the following solution: let two cars meet at the passing time t, then SA = VAT, sb = vb0t-12at2, SA = sb + △ s, The value of t can be obtained by combining the above formulas with the data. Do you think the student's solution is reasonable? If it is reasonable, please continue to calculate the time t; if not, please use your own method to calculate the correct result

A. Two cars B drive in the same direction on the straight road. Car a moves at a constant speed at the speed of VA = 20m / s, and car B moves at a constant deceleration straight line at the initial speed of vb0 = 24m / s, with the acceleration of a = 2m / S2. If car B is at △ s = 165m in front of car a, how long does it take for the two cars to meet? A student has the following solution: let two cars meet at the passing time t, then SA = VAT, sb = vb0t-12at2, SA = sb + △ s, The value of t can be obtained by combining the above formulas with the data. Do you think the student's solution is reasonable? If it is reasonable, please continue to calculate the time t; if not, please use your own method to calculate the correct result

In fact, when B car decelerates from uniform deceleration to stop, TB = 0 − V B0 − a = − 24 − 2S = 12s. Suppose the displacement of B car in this process is s, the solution is s = V b022a = 2422 × 2m = 144m. At this time, the displacement of a car is sa = VTB = 20 × 12m = 240m. Because SA ＜ S + 165m, a car has not caught up with B car. Then the distance a car passes is sa ′ = △ S + S = (165 + 144) M = 309m, so AB two cars meet Time t = SA ′ ν a = 30920s ≈ 15.5s. Answer: the solution is unreasonable, and the time to reach is 15.5s