If the maximum acceleration that passengers can endure in the subway train is 1.4m/s2. If the distance between two adjacent subway stations is 560m What is the shortest time for the subway train to travel between these two stations? The answer is 40,

If the maximum acceleration that passengers can endure in the subway train is 1.4m/s2. If the distance between two adjacent subway stations is 560m What is the shortest time for the subway train to travel between these two stations? The answer is 40,

Draw a V-T image to obtain an isosceles triangle with slopes of ± 1.4 and an area of 560
You can take the left half of this triangle (i.e. acceleration stage) s = (1 / 2) at ^ 2
Substitute the data s = 280m, a = 1.4m/s ^ 2 to get t = 20s ∵ symmetric ∵ t (total) = 2T = 40s

The maximum acceleration that passengers can bear in the subway train is 1.4m/s2. It is known that the distance between two necessary stops is 2240m Q: (1) assuming that there is no maximum limit, what is the minimum travel time between two stations? (2) If the maximum running speed of the train is 28m / s, what is the minimum running time between the two stations?

(1) 80s accelerates in half time and decelerates in half time. Assuming that the acceleration time is t, then 1 / 2 * 1.4 * T ^ 2 = 1 / 2 * 2240t = 40s, the total time is 2T = 80s (2) 100s. The calculation method of netizen zwb77777 is correct, but too careless. S1 = 1 / 2 * 20s * 28m / S = 280m instead of 140m. When the train accelerates to the maximum running speed

The maximum acceleration that passengers can endure in the subway train is 1.4m/s ^ 2. It is known that the distance between the two stations is 560m, so: (1) What is the minimum travel time between the two stations? (2) What is the maximum speed of the train between the two stations?

The minimum t is 40s
The maximum V is 101km / h
But this is unlikely
Generally, the subway has a certain track speed limit. At present, most of the speed limits are 80km / h

Trigonometric function: sin α+ cos α＝ 1 / 2, then sin α cos α= ( （sin α－ cos α）²= （

Square on both sides
sin ²α+ cos ²α+ 2sin α cos α= 1/4
1+2sin α cos α= 1/4
So sin α cos α=- 3/8
Original formula = sin ²α+ cos ²α- 2sin α cos α
=1+3/4
=7/4

Find indefinite integral ∫ (e ^ x-1) / (e ^ x + 1)

e^x=y
∫（e^x-1) / (e^x +1)dx
=∫（y-1) / (y +1)/ydy
=∫（2/(y+1)-1/y)dy
=2ln(y+1)-ln(y)
=2ln(e^x+1)-ln(e^x)
=2ln(e^x+1)-x

Find the indefinite integral of e ^ [(3x-1) ^ 1 / 2]

∫ e^√(3x-1) dx
Let √ (3x-1) = t
3x-1=t^2
x=(t^2+1)/3
dx=2t dt/3
‡ original formula = ∫ e ^ t * 2tdt / 3
=2/3∫e^t tdt
=2/3∫tde^t
=2/3 * te^t -2/3∫e^tdt
=2te^t /3 -2e^t /3 +C
=2√(3x-1) e^√(3x-1) / 3 - 2e^√(3x-1) / 3 +C