Uniform variable speed motion of physics in grade one of senior high school The motorcycle starts at rest, makes a uniform acceleration linear motion with the acceleration of A1 = 2m / S ^ 2, makes a uniform speed linear motion in the middle, and then makes a uniform deceleration linear motion with the acceleration of A2 = 4m / S ^ 2 until it stops. The total displacement is 1.8km. The duration t = 60s, and the solution is as follows (1) The maximum speed of a car in motion (2) When the acceleration of acceleration and deceleration process is the same as the original, and finally just stationary, the shortest time needed to walk this displacement and the maximum speed of getting off the car in this case (2) Because the time is the shortest, the time of uniform motion should be 0 Let v be the maximum speed v/2*v/2+v/4*v/2=1800 V = 40 root sign 3 The time is v / 2 * 1.5 = 30 root sign 3 The above is an answer from the Internet, I can't understand the formula of the second question, please explain!

Uniform variable speed motion of physics in grade one of senior high school The motorcycle starts at rest, makes a uniform acceleration linear motion with the acceleration of A1 = 2m / S ^ 2, makes a uniform speed linear motion in the middle, and then makes a uniform deceleration linear motion with the acceleration of A2 = 4m / S ^ 2 until it stops. The total displacement is 1.8km. The duration t = 60s, and the solution is as follows (1) The maximum speed of a car in motion (2) When the acceleration of acceleration and deceleration process is the same as the original, and finally just stationary, the shortest time needed to walk this displacement and the maximum speed of getting off the car in this case (2) Because the time is the shortest, the time of uniform motion should be 0 Let v be the maximum speed v/2*v/2+v/4*v/2=1800 V = 40 root sign 3 The time is v / 2 * 1.5 = 30 root sign 3 The above is an answer from the Internet, I can't understand the formula of the second question, please explain!

v/2*v/2+v/4*v/2=1800
Where V / 2 is the acceleration time and V / 2 is the average velocity
In the latter term, V / 4 is deceleration time and V / 2 is average speed
S = V, average T, no problem