The train advances at the speed of 20m / s. when passing station a, it needs to stop for 120s. The acceleration when entering the station is -0.5m/s2, and the acceleration when leaving the station is 0.4m/s2. After leaving the station, it still needs to move forward at the speed of 20m / s. how much longer does the whole train run because the train stops at station a?

The train advances at the speed of 20m / s. when passing station a, it needs to stop for 120s. The acceleration when entering the station is -0.5m/s2, and the acceleration when leaving the station is 0.4m/s2. After leaving the station, it still needs to move forward at the speed of 20m / s. how much longer does the whole train run because the train stops at station a?

It takes 40 seconds to slow down when entering the station and 50 seconds to accelerate when leaving the station, plus 120 seconds
The train walked 900 meters from deceleration to acceleration. It took 45 seconds to divide 900 meters by 20 meters per second,
Therefore, 120 plus 40 plus 50 minus 45 gets 165, so it takes 165 seconds more

As shown in the figure, the train is moving at a constant speed on a flat and straight railway at the speed of 54km / h. now it needs to stop temporarily at a station. If the dwell time at this station is 1min, the acceleration when the train enters the station is 0.3m/s2 and the acceleration when it leaves the station is 0.5m/s2, so (it is recommended to draw a motion scenario diagram) (1) What is the displacement of the train in the station? (2) What is the displacement of the train when it leaves the station? (3) How long was the train delayed by temporary parking at this station?

The train moves in a straight line with uniform deceleration with initial speed of V0 = 15m / s and acceleration of A1 = 0.3m/s2, then stands still, and finally moves in a straight line with uniform acceleration with initial speed of 0 and acceleration of A2 = 0.5m/s2 and final speed of V = 15m / s. The V-T image of train operation is shown in the figure. (1) the time required for the train to enter the station T1 = v0a1 = s =

The train moves along the straight track at the speed of V = 54km / h. for some reason, it needs to stop at a small station for to = 1min, and the acceleration when stopping at the station

Braking time t is not t0
Initial velocity V, final velocity VT = 0
Acceleration a = (final velocity initial velocity) / T
The conditions here are insufficient, and the acceleration is an indefinite value

There are two trains, a and B, with lengths of 200m and 300m respectively. It is known that the speed of car a is 54 km / h and that of car B is 10 m / s, (1) If two cars run in opposite directions, what is the time from car a catching up with car B to staggering with car B? (2) If two cars are traveling in the same direction, how long will it take from car a catching up with car B to staggering with car B?

V a = 54km / h = 15m / s, the distance when two vehicles pass by S = s a + S B = 200m + 300m = 500m;
（1）∵v=s
t，
‡ time when two vehicles drive in opposite directions:
t1=s
V A + v b = 500m
15m/s+10m/s=20s；
（2）∵v=s
t，
Time when two vehicles travel in the same direction:
t2=s
V a − v b = 500m
15m/s−10m/s=100s；
A: (1) if the two cars are driving in opposite directions, the time from car a catching up with car B to staggering with car B is 20s;
(2) If the two vehicles drive in the same direction, the time from vehicle a catching up with vehicle B to staggering with vehicle B is 100s

The train accelerates down the slope evenly. The speed at the top of the slope is 0.8m/s, the acceleration is 0.2m/s, and the time for the train to pass through the slope is 30s. Calculate the speed of this slope The length and the speed of the train at the end of 30s

V^2-0.8^2=2 × zero point two × X
V=0.8+0.2 × thirty
The solution is v = 6.8
X=114

The train starts to go downhill at the speed of 10m / s and gets an acceleration of 0.2m/s2 on the downhill road. When it reaches the bottom of the slope, the speed is 15m / s. calculate the time spent by the train passing through this slope road

T = V − V0 according to v = V0 + at
a＝15−10
0.2s＝25s．
A: it takes 25s for the train to pass this slope