A car starts to move at a constant speed in a straight line from a standstill. The distance from the start to 100 meters is that the speed increases by 10 meters per second, and the acceleration at the second 100 meters?

It should be "from rest to uniform acceleration linear motion"
The acceleration is constant. The acceleration at the second 100m is a = the acceleration at the first 100m
V^2=2aS
a=V^2/(2S)=10*10/(2*100)=0.5m/s^2

From the beginning to the first 100 meters, the speed of the car increased by 10 meters per second, passing the second 100 meters

（0+10）/2*t＝100
The first 100 driving time t = 20
at=10
The acceleration a = 0.5m/s
1/2aT^2=200
Driving two 100 times, sharing time t = 20 √ 2S
At this time, the speed: v = at = 10 √ 2m / S
Velocity increment V increment = V-0 = 10 √ 2m / S

A car starts to move in a straight line at a constant speed from a standstill. From the start to the first 100 meters, the speed increases by 10 meters per second,
What is the increase in speed when the car passes the second 100 meters?

From V1 ^ 2 = 2ax1, we can get 10 ^ 2 = 2A * 100
From V2 ^ 2 = 2ax2, we can get V2 ^ 2 = 2A * 200
From the above two formulas, we get V2 = 10, root sign 2
In the second 100 m, the increase of velocity is 10 knots, 2-10 = 4.14 M / s

A car starts to move at a constant speed in a straight line from a standstill. The distance from the start to 100 meters is that the speed increases by 10 meters per second, and the acceleration at the second 100 meters?