A physics problem about uniform variable speed linear motion in high school (about the proportional relationship in special laws) A train starts to move at a constant speed in a straight line from a standstill. One person stands in front of the first carriage to observe. The time for the first carriage to pass through him is 4 seconds, and the time for all trains to pass through him is 12 seconds. The distance between carriages is not recorded, and the length of each carriage is equal. It is required to (1) how many carriages in total; (2) How many cars passed this person in the last 4 seconds; How many seconds did the last three sections pass through this person? I hope it's best to use the proportional relationship in the special law, and attach the description and analysis, not just the formula and answer If the three questions are difficult to answer, you can answer the first one first

A physics problem about uniform variable speed linear motion in high school (about the proportional relationship in special laws) A train starts to move at a constant speed in a straight line from a standstill. One person stands in front of the first carriage to observe. The time for the first carriage to pass through him is 4 seconds, and the time for all trains to pass through him is 12 seconds. The distance between carriages is not recorded, and the length of each carriage is equal. It is required to (1) how many carriages in total; (2) How many cars passed this person in the last 4 seconds; How many seconds did the last three sections pass through this person? I hope it's best to use the proportional relationship in the special law, and attach the description and analysis, not just the formula and answer If the three questions are difficult to answer, you can answer the first one first

Is there a problem with the problem? Uniform motion? Is it uniform acceleration?
Set the length of each car as l, a total of N cars, and the acceleration as a;
l=1/2*a*4^2; nl=1/2*a*12^2; Divide the two formulas to get n = 9;
For the carriage passing in the last 4 seconds, you can first find the carriage passing in the first 8 seconds
N'l = 1 / 2 * a * 8 ^ 2, get n '= 4, so pass 9-4 = 5 cars in the last 4 seconds;
First find the passing time of the first 6 sections, 6L = 1 / 2 * a * T ^ 2; Get t = 4 √ 6;
So the last three sections are 12-4 √ 6

Uniform variable speed linear motion in Physics If an object falls from the top of the tower and the displacement passed in the last 1s before reaching the ground is 7 / 16 of the whole displacement, what is the height of the tower? (g takes 9.8m / S ^ 2) {keep one decimal place)

The displacement ratio per second in the falling process is 1:3:5:7:9:
Drop four seconds according to the topic
H=0.5gt^2=8*9.8=78.4m

A police car on duty stopped on the highway. When the police found that the truck running at a constant speed of V = 8m / s next to him had violations, they decided to stop it. After 2.5s, the police car started and drove out at a constant acceleration of a = 2m / S2 to maintain a constant acceleration. The maximum speed that can be achieved is 180km / h (1) How long does it take the police to catch up with the illegal truck? (2) What is the maximum distance between the two workshops before the police car catches up with the truck?

(1) 180km / h = 50M / s, the time when the police car reaches the maximum speed is T1 = VMA = 502s = 25s, and the displacement of the police car is x = 12at21 = 12 × two × 252 ＝ 625M, the displacement of freight car is: x2 ′ = V (t ′ + 2.5) = 8 × 27.5m = 220m, it is known that the police car has caught up with the truck before the maximum speed. Police officers are set to start

Physical acceleration and uniform velocity motion formula Better have it all! I'm waiting to use it^-^ Bye, drag! The more complete, the better! Sheila!

x=vt+1/2at^2
V=v+at
V^2-v^2=2as

A physical problem of the relationship between variable speed motion and acceleration The motion of an object on a straight line gives the positive and negative initial velocity and acceleration. The correct description of motion in the following pairs is () A．v0>0,a0,a

B. Remember that if the speed is in the same direction as the acceleration, then it is acceleration, no matter the acceleration becomes larger or smaller. On the contrary, it is deceleration

On the uniform accelerated linear motion of physical objects The train starts to move in a straight line with uniform acceleration a from a standstill, and then runs at uniform deceleration of a / 3 acceleration along the original direction to C. for example, the distance between a and C is 18km, and the train has been running for 20min. Calculate the distance between a and B and the value of acceleration a

Let the distance between Party A and Party B S1 time T1 between Party B and Party C S2 time T2 between Party A and Party C s total time t reach Party B, and the speed is V (International System of units)
Then s = S1 + S2......... 1
S1= 0.5at1^2…………2
S2= 0.5*（1/3）at2^2……………3
t=t1+t2………………4
t1=V/a………………5
t2=V/(a/3)…………6
Substitute the data into equation 1 2 3 4 5 6
S1=4500m
a=0.1m/(s^2)