1. The original speed of the locomotive is 36km / h, and the acceleration is 0.2m/s on a section of downhill road. When the locomotive runs to the end of the downhill, the speed increases to 54km / h. find out the time for the locomotive to pass this section of downhill road 2. The train should slow down in advance when passing through bridges and tunnels. A train running at a speed of 72km / h decelerates at a constant speed when entering a stone arch bridge. It decelerates for 2min, and the acceleration is 0.1m/s ^ 2. What is the speed of the train after decelerating?

1. The original speed of the locomotive is 36km / h, and the acceleration is 0.2m/s on a section of downhill road. When the locomotive runs to the end of the downhill, the speed increases to 54km / h. find out the time for the locomotive to pass this section of downhill road 2. The train should slow down in advance when passing through bridges and tunnels. A train running at a speed of 72km / h decelerates at a constant speed when entering a stone arch bridge. It decelerates for 2min, and the acceleration is 0.1m/s ^ 2. What is the speed of the train after decelerating?

1、25s
vt=vo+at
15=10+0.2*t t=25
2、vt=vo+at
vt=20m/s+0.1*120=32m/s

Find the speed
On a smooth horizontal plane, the ball a with mass m moves towards the ball B with mass 3M at a speed v
After the collision, the velocity of ball a is v. / 4

Do it with the momentum theorem!
Take the direction of a ball as the positive direction!
m*V+0=3m*V(B)-m*V/4
4m*V=12m*V(B)-m*V
5m*V=12m*V(B)
V(B)=5V/12

With the rapid development of China's expressway network, in order to ensure safety, the necessary distance should be kept between the vehicles on the expressway. It is known that the maximum speed limit of a certain expressway is v = 120km / h. suppose that the vehicle in front stops suddenly, and the driver behind operates the brake when he finds this situation, and the time (i.e. reaction time) that the vehicle begins to decelerate is t = 0.05s, When braking, the acceleration of the car is m / S2. Try to calculate the distance X between cars on the highway at least?

At least, it should be assumed that the brake of the car in front stops, and the displacement generated from the brake to the stop just reaches the position of the car in front. From the beginning of the brake to the stop, it is calculated by two kinds of motion. The reaction time period is uniform motion, and the brake starts to decelerate with the acceleration of a (where a is negative) to the stop, v = 120km / h = 100 / 3m / s
The first stage of uniform motion: S1 = v0.05 = 5 / 3m / s
The second stage of uniform deceleration: V end = V + at; t = - V / A
S2=vt+1/2at^2=-v^2/a-1/2v^2/a=-3/2v^2/a
Because acceleration is not given, it can only be replaced by symbol. I hope it can help you

Such a sudden factor must be considered in the design of runway length: when the aircraft taxis on the runway, it accelerates at a constant acceleration of a = 4.0 m / S2. When the speed reaches 90 m / s, it can take off. But at this time, it can not take off for some reason, so it should brake immediately, and the maximum acceleration generated by braking is 5 m / S2?

Therefore, the runway length should be at least s = S1 + S2 = 1822.5m