A car traveling along a straight highway with uniform acceleration, passing two electric poles 50m apart by the roadside, shares 5S. When it passes the second electric pole, its speed is 15m / s, and when it passes the first electric pole, its speed is () A. 2m/s B. 10m/s C. 2.5m/s D. 5m/s

A car traveling along a straight highway with uniform acceleration, passing two electric poles 50m apart by the roadside, shares 5S. When it passes the second electric pole, its speed is 15m / s, and when it passes the first electric pole, its speed is () A. 2m/s B. 10m/s C. 2.5m/s D. 5m/s

The average speed of an object passing through two poles is:
.
v=x
t      ①
Since the object moves in a straight line with uniform acceleration, there are:
.
v=v1+v2
two     ②
The data obtained by substituting the two formulas: V1 = 5m / s, that is, the speed passing through the first electric pole is 5m / s, so ABC is wrong and D is correct
Therefore, D

Physics uniform speed linear motion problem You go to the railway station to see your friend off. Your friend just sits in the first carriage No. 1. The train accelerates slowly and evenly from your side. If the first carriage passes by you, sometimes 2S, and the whole train takes 8s, how many carriages does the whole train have? How long will it take you for the last carriage to pass by?

S = 1 / 2at1 ^ 2 so: S = 1 / 2A (2) ^ 2
Ns = 12 / ATN ^ 2 so: SN = 1 / 2A (8) ^ 2
It can be obtained by combining the above two types: n = 16 (with 16 cars)
s=1/2a*2^2
15s=1/2a(t1)^2
16s=1/2a(t2)^2
Combined with the above three formulas, t2-t1 = 8-60 ^ 1 / 2 = 0.254s

The problem of uniform speed linear motion in Physics 1. It is reported in the news that "the maximum speed of the maglev train is 430km / h, and the total length of the maglev track from Zhangjiang to Shanghai Pudong International Airport is 29.863km, and the whole process only takes 7min". Assuming that the acceleration of starting and braking is equal, the driving time at the maximum speed is ____, The acceleration is ___ 2. Starting from standstill, the car makes uniform acceleration with A1, then makes uniform deceleration with A2 after a period of time, and finally stands still. It has advanced a total distance of L, so the total time of car movement is ____

16．66．7 m／s 20 rain
Maximum train speed VM = 430 km / h ≈ 120 m / s,
1. The accelerations in acceleration phase and deceleration phase of the train are equal, so
The displacement of acceleration section and deceleration section shall be equal, set as x1, and the time taken shall be equal, set as T1,
Let the acceleration be a m / S ^ 2
Total length x = x0 + 2x1 = 29.863 km
Total time t = t0 + 2t1 = 7 × 60 = 420 s
v^2 = 120^2 = 2 * a * x1
2 * X1 + 120 * t0 = 29863, so a = 0.589; t1 = 203.75 s； t0 = 12.5 s
Or x = VM * t0 + 2 * 0.5 * a * T1 ^ 2 = 29.863 km
t = t0 + 2t1 = 7 × 60 = 420 s
v = a * t1
Maximum speed = acceleration * T1, so acceleration = 120 / T1 M / S ^ 2
2 answer: √ [2L (a1 + A2) / (A1 * A2)]
Let the car start from point a, reach point B through time T1, and then reach point C through T2
Let the speed of the vehicle at point B be VB, the average speed of the vehicle moving from point a to point B VAB = VB / 2, and the displacement SAB = VB * T1
The average speed VBC = VB / 2 and displacement SBC = VB * T2 during the movement of the vehicle from B to C
Then l = SAB + SBC = VB * (T1 + T2) / 2 ①
And T1 = VB / A1; t2 = VB/a2 ；
So t = T1 + T2 = VB / A1 + VB / A2 = VB [(a1 + A2) / (A1 * A2)] ②
① ② simultaneous, eliminate VB, and get t = √ [2L * (a1 + A2) / (A1 * A2)]

Application of uniform speed change linear motion law in physics problems When a particle moves in a uniformly accelerated straight line with an initial velocity of zero, its displacement in the first second is 2 meters. What is the displacement of the particle in the 10s? How long does it take for a particle to pass through the third 2S? Write down the steps. Thank you

In the first second: S = (1 / 2) at ^ 2A = 2S / T ^ 2 = 2 * 2 / 1 = 4m / S ^ 2 displacement in the 10s s s = front 10s position Ρ - front 9s position Ρ = 0.5 * 4 * (10 * 10) -0.5 * 4 * (9 * 9) = 38m time to pass through the third 2m t = time to pass through the first 6m T1 - time to pass through the first 4m T2S = (1 / 2) at ^ 2T = root sign (2S / a) t = root sign (2 * 6

On the law of linear motion with uniform speed change The car is moving on the straight road at the speed of 10m / s. there is a bicycle at so in front of it. It is moving in the same direction at the speed of 4m / s. The car immediately turns off the accelerator and makes a uniform linear motion of a = - 6m / S2. If the car just can't touch the bicycle, the size of so is () A.9.67m B.3.33m C.3m D.7m The initial speed is 10m / s and the final speed is 4m / s. VT = VO at calculates the time t = 1s. In one second, the car travels VOT + (1 / 2) at * t = = 7m minus the 4 meters traveled by the bicycle in one second, and the result is 3 meters

Choose C. when the speed of the car is the same as that of the bicycle, the car just can't touch the bicycle. It takes one second to reduce the speed of the car to 4m / s, and the journey is 7m, while the bicycle advances 4m in one second, the difference between the two is 3M, so the correct answer is C

High school physics problems (about the law of uniform speed linear motion) should be analyzed A car was driving at a constant speed of 10m / s. when it was 24.5m away from the intersection in front, the red light was on. The driver immediately drove at 2m / s ² The acceleration of braking, ask whether you will break the rules due to running the red light after braking for 6S?

It can be seen from the meaning of the question: the initial speed of the car is V0 = 10m / s, a = 2m / s, then the time of complete standstill is: T = V0 / a = 10 / 2 = 5S
Therefore, the displacement in 5 seconds is:
s=v0^2/(2a)=25 m>24.5m
That is, the car passed the intersection and ran the red light