Brake the train before entering the station. It is known that the train speed before braking is 60km / h and the braking acceleration is 0.8m/s2. Calculate the train speed 15s and 30s after braking

Brake the train before entering the station. It is known that the train speed before braking is 60km / h and the braking acceleration is 0.8m/s2. Calculate the train speed 15s and 30s after braking

60km / h = 16.67m/s, time required for train speed to decrease to zero t0 = 0 − v0
a＝0−16.7
0.8=20.9s．
Then the speed at the end of 15s after braking v = V0 + AT1 = 16.67-0.8 × 15=4.67m/s．
The speed at the end of 30s is zero
Answer: the speed at the end of 15s after braking is 4.67m/s, and the speed at the end of 30s is zero

For a train with uniform acceleration, the speed increases from 10m / s to 20m / s within 40s to calculate the acceleration of the train. When the car brakes in an emergency, the speed decreases from 10m / s to zero within 2s to calculate the acceleration of the car?

The acceleration of the train is: A1 = V2 − v1
t＝20−10
40＝0.25m/s2；
Acceleration when braking: A2 = 0 − v0
t＝−10
2＝−5m/s2
Answer: the train acceleration is 0.25m/s2;   The acceleration of the vehicle is 5m / S2

A train starts from a standstill and makes a uniform acceleration linear motion with an acceleration of 1m / S ^ 2. After 20s, it makes a uniform linear motion with its final speed. After a period of time, it starts to make a uniform deceleration linear motion with an acceleration of 2m / S ^ 2 and finally stops. The displacement of the train in all movements is 10km, and the time for the train to make a uniform linear motion is calculated

S = 200m during uniform acceleration
S = 100m during uniform deceleration
Therefore, the uniform process t = s ÷ v = (10000-300) ÷ 20 = 485s
Therefore, t = 20 + 485 + 10 = 515 seconds

The car is traveling at the speed of 54km / h, and it needs to slow down and stop at a certain place. It is set that the car brakes at the acceleration of - 3m / S ^ 2. How far has it gone from start to stop

V end = 0, V beginning = 15, a = - 3
V end = V beginning + at 0 = 15 + (- 3) t
t=5
S = t * (V end + V beginning) / 2 = 5 * 7.5 = 37.5m
Or S = V initial T + (1 / 2) at ^ 2 = 15 * 5 - (3 / 2) * 25 = 37.5m
Pay attention to converting 54 km / h into 15 m / s

The vehicle speed is 1.5m / s, the vehicle stops for 2min for some reason, and the vehicle acceleration during braking is 0.3m per quadratic second When starting, the acceleration is 0.5m, and the specific steps should be taken to calculate the delay time of the car due to parking every quadratic second!

The delay time of the vehicle for some reason is the time period from braking to acceleration to 1.5m/s, which is divided into three parts
1. Braking period: VT = V0 + at, 0 = 1.5-0.3t, t = 5S
2. Stop halfway for 2min
3. Start acceleration section: VT = V0 + at, 1.5 = 0 + 0.5T, t = 3S
Delay time due to parking = 2min and 8s

The car runs at the speed of 54 kilometers per hour. For some reason, it stops for 2 minutes. The acceleration of braking and starting are 0.3 square meters per second and 0.5 square meters per second respectively. Try to find the time delayed by the car

Isn't the acceleration unit right? It's 0.3m/s square and 0.5m/s square
Speed v = 54km / h = 15m / s, braking time T1 = V / 0.3 = 50s
Acceleration time T2 = V / 0.5 = 30s
The total delay time is equal to 50 + 30 + 120 = 200s, i.e. 3 minutes and 20 seconds