The car stops suddenly after driving for 5 minutes at a speed of 20 m / s. if the braking process is a straight-line motion with variable speed, the acceleration is the square of 5 m / s, What is the distance of the car in the last second?

The car stops suddenly after driving for 5 minutes at a speed of 20 m / s. if the braking process is a straight-line motion with variable speed, the acceleration is the square of 5 m / s, What is the distance of the car in the last second?

Vt-V0=at
0-20=-5t
T = 4 seconds
s=v0t+att/2
The last 1s forward distance S = s4-s3 = 20 * 4-5 * 4 * 4 / 2 - (20 * 3-5 * 3 * 3 / 2)
=40-37.5
=2.5m

When the car is driving on a straight road, it first accelerates at a constant acceleration of 0.5m/s and 178; from a standstill. When the car's speed reaches 10m / s, it drives at a constant speed for 15s. Then it brakes at an acceleration of 2m / s and 178; in order to wait for the pursuers. How far is the car going? How long is it going?

When the vehicle speed reaches 10m / S
Time t = 10 / 0.5 = 20s
Displacement △ x = 1 / 2x0.5x400 = 100m
Displacement at constant speed △ x = 10x15 = 150m
Braking time 10 / 2 = 5S
The displacement in 5S is △ x = 10x5-1 / 2x2x25 = 25m
The car advanced 100 + 150 + 25 = 275m
Forward 20 + 15 + 5 = 40s

The speed of the car before braking is 20 meters per second, and the acceleration obtained by braking is 2 meters per square second
Find (1) the taxi distance within 20 seconds after the start of braking
(2) The time from the beginning of braking to the displacement of 30 meters
(3) Taxi distance in 2.5 seconds before standstill

1）v^2=2as,s=v^2/(2a)=20^2/(2*2)=100m
2) Using the area enclosed by V-T lines, let t:
(v0+v0-at)*t/2=s1,(20+20-2t)*t/2=30,
Because t (max) = V / a = 20 / 2 = 10s,
T = 10-radical 70
3) We can also use V-T line to calculate v = at = 2 * 2.5 = 5m / s,
s=(5+0)*2.5/2=6.25m