The car is moving in a straight line at a speed of 10 meters per second on a straight road. When you find something in front of you and brake, the acceleration you get is as follows 2 meters per second (1) The speed of the car at the end of three seconds (2) The speed at the end of five seconds (3) The speed at the end of 10 seconds

The car is moving in a straight line at a speed of 10 meters per second on a straight road. When you find something in front of you and brake, the acceleration you get is as follows 2 meters per second (1) The speed of the car at the end of three seconds (2) The speed at the end of five seconds (3) The speed at the end of 10 seconds

（1）v3=v0+at=10-2*3=4m/s
(2) Car stop sharing time t = 10 / 2 = 5S
v5=v0+at=10-2*5=0
(3) The speed is zero at the end of 5 seconds and zero at the end of 10 seconds

The car moves in a straight line at a speed of 10 m / s on a straight road. When it finds something in front of it, it brakes. It is known that the acceleration after braking is 2 m / S ^ 2
Then (1) if the reaction time of people is not considered, what is the total braking time in order to avoid traffic accidents?
(2) How far does the car coast during braking?

(1) The total braking time of a car is the time required for the speed of the car to decrease from 10m / s to 0 with an acceleration of - 2m / S ^ 2
So t = △ V / a = (0m / s-10m / s) / - 2m / S ^ 2 = 5S
(2) This small question is to find the displacement of the car in the process of deceleration. I offer you two methods, one is simple, the other is conventional
① General: 2aX = V & # 178; - V0 & # 178;, so we get x = 25m
② Simple: because the average velocity is equal to (V + V0) / 2, and the displacement is equal to the average velocity times time, the average velocity is 5m / s, and the time is 5S, so the displacement is 25m
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When the car moves in a straight line with a uniform initial speed of 10 m / s on a straight road, it brakes when there is something in front of it, and the acceleration is 2 m / S & sup2
(1) What is the speed of the car in three seconds?
(2) What's the speed in five seconds?
(3) What is the speed after 10 seconds?

1)
Vo=10m/s,a=-2m/s^2
The velocity after T1 = 3 seconds is
V1=Vo+6t1=10+(-2)*3=4m/s
2)
When T2 = 5 seconds, the velocity is
V2=Vo+a*t2=10+9-2)*5=0
3)
The car stops 5 seconds after starting to brake, and the speed is 0 after 10 seconds

When the car moves in a straight line at a speed of 10m / s on a straight road, it brakes immediately when it finds something in front of it, and the acceleration is 2m / s * 2
(1) What is the speed of the car after 3S?
(2) What is the speed of the car after 5S?
(3) What is the speed of the car after 10s?

The results show that when the vehicle moves at a constant deceleration, V0 = 10m / s, a = - 2m / s and#178;
It can be calculated directly according to v = V0 + at
Just pay attention: the time t '= 5S when the car speed is reduced to zero, that is, the car speed has been reduced to zero at the end of 5S,
Therefore, the speed is zero after 10 seconds
If necessary, I'll write the process for you