The train runs along the track straight line from ground a to ground B. the distance between B is s. The train starts from ground a and accelerates uniformly. The linear motion acceleration is A1, then decelerates uniformly. The linear motion acceleration is A2. When it arrives at B, it just stands still. Calculate the time t for the train from a to B

The train runs along the track straight line from ground a to ground B. the distance between B is s. The train starts from ground a and accelerates uniformly. The linear motion acceleration is A1, then decelerates uniformly. The linear motion acceleration is A2. When it arrives at B, it just stands still. Calculate the time t for the train from a to B

Find the shortest time t for the train from a to B, then the train accelerates first and then decelerates. There is no constant speed process (it can be proved by V-T diagram)
So a1t1 = a2t2
t=t1+t2=t1+a1t1/a2=t1(1+a1/a2)
t1=t/(1+a1/a2)
Overall average speed v = a1t1 / 2 = a2t2 / 2
t=d/(a1t1/2)
=2d/a1t1==2d/a1[t/(1+a1/a2)]

The object drives from ground a to ground B in a straight line along the track, and the distance between a and B is D. the train starts from ground a and makes a uniform acceleration linear motion, with the acceleration of A1. The train makes a uniform deceleration linear motion in the last stage, with the acceleration of A2. When it reaches B, it just stands still. The train can also make a uniform motion during the driving process, so as to calculate the shortest time t for the train from a to B

In order to minimize the time for the train from a to B, the train must first accelerate uniformly and then decelerate uniformly. Suppose that the time for the object to accelerate uniformly and decelerate uniformly in a straight line is T1 and T2 respectively, and the maximum speed is v. then: v = a1t1, v = a2t2 get a1t1 = a2t2, which is transformed from mathematical knowledge to: t2t1 = A1A2

If someone travels for a week, the sum of the days of the week is 91, then the last day of the day is

(91+1+2+3+4+5+6)/7=112/7=16

Xiaobin travels for a week on vacation. The sum of the days of the week is 84. When did Xiaobin go home?

84/7=12
So this week is 9, 10, 11, 12, 13, 14, 15
Back on the 15th

Xiao bin travels for a week during his holiday. The sum of the days of the week is 84. When did Xiao bin return home

84/7=12
So the date of going out is 9,10,11,12,13,14,15
All come back on the 15th

(Application of one yuan to the power) Xiaobin went out for a week in August this year. The sum of the days in this week is 84. Do you know when Xiaobin came home?

Suppose the first day of the week is x, then the second day is x + 1, the third day is x + 2,... And so on, and the last day is x + 6
x+x+1+x+2+x+3+x+4+x+5+x+6=84
7x+21=84
7x=63
The solution is x = 9
So x + 6 = 15
A: Xiao bin came home on the 15th