A physics question about linear motion physics questions about linear motion. To detail the process

The former one can't.

The latter can.

It's just a straight line problem of uniform force, and the following solution was conceived by myself in my freshman year of high school and studied with some college classmates, according to calculus, we can know that v=3t 2+1 can be obtained, a=6t, and v0=1, so let the amount of variable force be h, h=6, according to calculus x'=v=1 2ht 2+at+v0, so x=1 6ht 3+1 2at 2+v0t+x1, so xb+5=xa=5t 2+5=t 3+t, you can get, t=5, so, xa=xb=5*5*5+5=130m

1) The distance from the motorcycle after the vehicle is started.

d = v mo (t anti + t hair) = 8m s (

After the police car starts, it will take time for T2 to catch up.

Mot2+dt2 -8t2-20=0

t2-10)(t2+2)=0

t2=10s, from the start of the police car, 10s to catch up, and from the motorcycle to pass, to catch up 2) when the speed is the same, the distance is the largest, it takes time.

t1=v mol a = 8m s 2m s =4s

Solution:1The perpetrator of the motorbike is gone.

When the traffic police catch up, the traffic police walk the same distance as the perpetrator. i.e.: s1=s1

Let the time be t, so 24 + 10t = squared). a is the acceleration. Solution t=12 seconds.

Second question. Let the maximum distance be l, and the time is tthen l=24+squared).

To make l the maximum, i.e., find the vertices of this parabolic equation. That is, there is a maximum value when t=5. So the police car drove for 5s

b There is smoke coming out of the chimney, it can be seen that the current wind direction is from right to left, car A is stationary or moving to the right or moving slowly to the left, the direction of the flag will be as shown in the figure, and car B must be moving to the left.

Because the smoke of the house drifts to the left, the wind blows to the left.

So there are two possibilities for the state of motion of the car 1. Movement to the right 2. Rest.

Because the flag of car B is floating to the right, car B can only move to the left, and the speed is greater than the speed of the wind, so D is chosen

First, find the critical time t=2t, t is the time required to decelerate to 10m s, and t is the total time of deceleration.

Because it is a uniform deceleration, the average speed is 15m s, 15m s*t-10m s*t=s=100m, so t=20s, so t=40s

Therefore, in order not to cause A and B to collide, the deceleration time of car A should not exceed 40s

You didn't give the acceleration of the deceleration. sorry。

The train does a uniform deceleration motion, and the acceleration a is constant, so the time it takes to reduce the speed to half of the initial velocity is also half of the total time, t1=t2=20s, and the system of equations: and. where v = 1 2v0.

Then solve v0 and a, and I believe you will be able to do it later.

Because it is a uniform deceleration, the time taken when x1 = 30 m t1 = t2 = 20s t = 40s

then (v + v = 2m s.)

x=x1+x2=x1+

6m s, 40s, 40m, using the instantaneous velocity at the middle moment in the uniform deceleration motion is the average of the sum of the initial velocity and the final velocity.

In the first s, it accelerated from a standstill to 1 m s, and it moved.

s=1/2at＾2=

In the second second, it decelerates from 1m s to zero and moves again.

And so on, the displacement in every second is.

Therefore, the total displacement of this particle after 100s is 50m

The displacement of the object at the 1st s.

s1 = 1 2at 2 = 1 2 * 1 * 1 2 = 1 2 m when the velocity is 0 at the end of the second second

The motion of the 2nd S is a uniform deceleration and linear motion, and the displacement at this time.

s2=1 2at 2=1 2*1*1 2=1 2 m and so on.

The total displacement of 100s is.

1/2m×100=50m

The acceleration in the first second is 1m s, the displacement is 1 2 at 2=, the acceleration in the second second is -1, and the deceleration motion is made, the velocity at the end of the second second is 0, and the displacement in the second is the same, and so on, the velocity is added to 1 at the end of every odd second, and the deceleration is 0 at the end of every even second, and the displacement is every two seconds, so after 100 seconds, the velocity is 0m s, and the displacement is 50m.

Divided into 50 small segments to analyze, that is, every 2 seconds is 1 segment, the sum of the front and rear forces is zero, you can see the whole as a uniform motion, and it should be very simple to analyze the movement in 2 seconds, right? First accelerate, then decelerate, at the beginning of the 3rd second, the speed is 0, the displacement is 2 meters, the details are not explained, and then 2m 50 segments = 100 meters. Are you sure your answer is correct?

Oh, I miscalculated, I forgot to multiply the 1 2 before at 2, I moved 1 meter in 2 seconds, and the total distance was 50 meters, yes, I'm sorry, it's been a long time, the formulas are rusty, hehe.

As a V-T diagram, we can see that the image is 50 isosceles triangles, and the sum of their areas is found.

s=1*50=50

Average speed land concessions.

Neutral velocity [velocity at the midpoint of a displacement time].

B is a point between AC, then half of the early training room at AB plus half of the BC time = half of the total time = T 2

The velocity at the time point in AB is V, BC is 2V, the amount of change is V, and the time taken is T 2, so the acceleration = V T 2 = 2V Midfield T

aDistance traveled = (3t 2+1)dt=t 3+t+c-0-0-c=

t 3 + t, b traveled distance = 5t 2, then there is.

t 3 + t = 5t 2 + 5, t = 5s before solving the celery ant, and the distance traveled by a is 130m