After braking at the same time, the two vehicles of a and B are moving in a straight line with uniform deceleration. A stops for 3 seconds and advances for 36 meters. B stops for 1.5 & nbsp; s. The forward distance of B is () A. 9 mB. 18 mC. 36 mD. 27 m

After braking at the same time, the two vehicles of a and B are moving in a straight line with uniform deceleration. A stops for 3 seconds and advances for 36 meters. B stops for 1.5 & nbsp; s. The forward distance of B is () A. 9 mB. 18 mC. 36 mD. 27 m

After braking, a car moving at a speed of 18m / s will decelerate evenly and advance 36m in 3S to calculate the acceleration of the car,

First calculate the average velocity, vmean = s / T = 12m / s,
For uniform velocity, if vmean = (VT + V0) / 2, VT = 6m / s can be obtained;
a=（vt-v0）/t=-4m/s^2.

If the speed of car a and car B is equal, and the car decelerates evenly after braking, then the distance of car B is ()
A. 9 mB. 18 mC. 36 mD. 72 m

Let the initial velocity of a and B be v0. According to the inference of the average velocity, the displacement of a is X1 = v02t1, that is, 18 = V02 × 3. The displacement of B is x2 = v02t2 = V02 × 1.5. The simultaneous solution of the two equations gives x2 = 9m. So a is correct and B, C and D are wrong

The two cars of a and B brake at the same time with the same speed, and they all decelerate evenly. A stops for 3 seconds and moves forward for 36 meters, B stops for 1.5 seconds and B moves forward for a long distance
It's not 18, is it

It should be 18. Calculate the acceleration 8m / S ^ 2 from a
Then we can use s = 0.5at ^ 2 to calculate the distance of B. here we can use this formula. Don't think I used it wrong, because the final velocity is 0, and the process is equivalent to the uniform acceleration motion with the initial velocity of 0