It is known that the maximum speed of an expressway is v = 120km / h. assuming that the vehicle in front stops suddenly, the driver of the vehicle behind finds this situation, operates the brake and arrives at the vehicle The time to start deceleration (i.e. reaction time) t = 0.50s, the magnitude of vehicle resistance when braking is 0.40 times of vehicle gravity, what is the minimum distance s between vehicles on the expressway? (gravity acceleration g is taken as 10m / S2) If f = am, how does f / a come from

It is known that the maximum speed of an expressway is v = 120km / h. assuming that the vehicle in front stops suddenly, the driver of the vehicle behind finds this situation, operates the brake and arrives at the vehicle The time to start deceleration (i.e. reaction time) t = 0.50s, the magnitude of vehicle resistance when braking is 0.40 times of vehicle gravity, what is the minimum distance s between vehicles on the expressway? (gravity acceleration g is taken as 10m / S2) If f = am, how does f / a come from

When braking, there is resistance in the horizontal direction
Ff=0.4mg
a=Ff/m=0.4g
v=120km/h=100/3m/s
2ax1=v^2
x1=1250/9m
x2=v*t=100/3*0.5m=50/3m
s=x1+x2=1250/9+50/3=1400/9m≈155.56m

There are signs everywhere on the expressway to remind drivers to keep a reasonable distance. Try to estimate the distance according to the following data: the car is driving on the expressway at the speed of 120km / h, and the car is in front of the road
When the car stops on the road due to a fault, the driver takes 0.6s to brake immediately when he finds out the situation. After the emergency braking, the acceleration of the car sliding forward is 6m / S2, which can be regarded as uniform deceleration movement?

According to the design standard of highway engineering, the stopping sight distance of car with design speed of 120 km is 210 m, and that of truck is 245 m (the result is related to the eye height of car)
Stopping sight distance refers to the shortest driving distance from the time when the driver sees the obstacle in front to the time when the driver stops safely in front of the obstacle
The stopping sight distance is composed of three parts: the driving distance S1 in the driver's reaction time, the driving distance S2 (braking distance) from the beginning of braking to the complete stop of the vehicle, and the safety distance S0 (5-10m). It is calculated by the following formula: St = S1 + S2 + S0 = (u1t / 3.6) + U12 / (254 φ 2) + S0, where: T - driver's reaction time, Take 2.5s; φ 2 - longitudinal friction coefficient between road surface and tire, which varies with tire, road surface, braking and other conditions, and calculate stopping sight distance according to wet road condition; U1 - driving speed

When a car started from a road sign and drove south for 1000m, the driver suddenly remembered that a tool had been left 3m to the north of the road sign, so he went back along the original road
After finding the tool, continue to drive south for 600m, and stop there to have a rest. It is trying to show several special positions on the coordinate axis, and find out the displacement and distance

1000、1003、603

On the highway, some drivers in order to reduce operating costs, wantonly overload, bring great harm
And the answer has to be calculated

(1) G total = (M vehicle + m cargo) g = (2000kg + 10000kg) × 10N / kg = 1.2 × 105np = f / S = g total / S = 1.2 × 105N △ 6 × 0.02m2 = 10 × 105Pa, 10 × 105Pa ＞ 7 × 105Pa, exceeding the industry standard (2) g not exceeding the standard = f not exceeding the standard = P not exceeding the standard? S = 7 × 105Pa × 0.12m2 = 0.84 ×