High School Physics Linear Motion Problems, Senior 1 Physics Linear Motion

The displacement sn of the nth second is equal to the displacement of the previous n seconds minus the displacement of the previous (n-1) seconds.

That is, sn=1 2g(n2-(n-1) 2)a, bring n=1 into sn=5m

b brings n=2 to get sn=15m

c brings n=3 to get sn=25m

d, bring n=4 into sn=35m

So choose D

b: The displacement of an object in 2 seconds is: the displacement of the previous 2 seconds minus the displacement of the previous 1 second.

s1=1/2gt1^2=5m

s2=1/2gt2^2=

The displacement of the object in 2 seconds is: s2-s1=15mc: similarly: s3= 1 2gt3 2=

The displacement of the object in 3 seconds is: s3-s2=25md: s4= 1 2gt4 2=

The displacement of the object in 4 seconds is: s4-s3 = 35m

Landlord, please read the title carefully, it is about the displacement in a few seconds, understand? The displacement in the second is the total displacement of the object when it moves for 2 seconds minus the displacement in the first second, the answer is d, you should understand it by now, since you can say that bc is not right, I believe you will do it.

The displacement in the 4th second should be 35 meters.

Only d is wrong, the rest are right.

x=1/2*gt²

a: When t=1, x=5

B: When t=2, x=20, so δx=20-5=15c: when t=3, x=45, so δx=45-15-5=25d: when t=4, x=80, so δx=80-25-15-5=35

An important conclusion of linear motion with uniform variable speed:

Uniform acceleration from a standstill (from a standstill, of course, it cannot be evenly decelerated!) ), the displacement ratio for the same period of time in a row is:

2n-1) First two seconds: Second two seconds: Third two seconds: Fourth two seconds 1:3:5:7

That means that the fourth displacement in two seconds is 3 7

21 meters. So where did this conclusion come about? By formula: s

1 2at 2 comes. Hey? Where did the "v0t" go? The initial velocity is 0, and v0t is 0, so there is no need to write it here.

As long as it is a continuous period of the same length, there is the above ratio relationship.

Proof : Let these time periods be continuous, and the length of time is t, then.

Displacement of the first time t: s1

1/2at^2

Displacement of the first two t times: s2

1/2a(2t)^2

Displacement of the first three t times of 4s1: s3

1/2a(3t)^2

The first four t times of 9s1 .........16s1

Displacement of the first n-1 t time: (n-1) 2s1

Displacement of the first n t times: n 2s1

That is, it is all square growth.

So, what is the displacement of the second t-time? Of course, the displacement of the first two t-times, minus the displacement of the first t-time! That is, (4-1)s1

What is the displacement of 3s1 at the third t time? Of course, the first three minus the first two! That's 5s1

What is the displacement at time nth t? Of course, the first n, minus the first n-1, that is. n^2s1

n-1)^2s1

2n-1) s1 up.

Therefore, the ratio of the displacement passed by a uniform acceleration linear motion with an initial velocity of 0 in a continuous, equal-length period of time is.

2n-1)

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

The instantaneous velocity of an object moving in a straight line with uniform acceleration at the middle moment of a certain period of time = the average velocity during this time.

The time is counted from the time of entering the first displacement.

t=(1 2)t1 is the middle moment of t1, and the instantaneous velocity at this time v=average velocity in t1=s t1

t'=t1+(1 2)t2 is the intermediate moment of t2, and the instantaneous velocity v at this time'=average velocity within t2=s t2

a=(v'-v)/(t'-t)=2s(t1-t2)/[t1*t2*(t1+t2)]

Answer: If the displacement is 2s, then the general displacement is s

The average velocity of the first half of the displacement is v1 = 4m s

The time is t1=s v1=s 4

The average velocity of the latter half of the displacement is v2=8m s

The time is t2=s v2=s 8

So the total time t=t1+t2

The average velocity of the whole process is v=2s t substituting the above equation.

v=16/3m/s

vt=v0+at...From point A to point D is 3 10 seconds. Because I didn't flash 10 times a second...

The starting speed is 0. Because s=v0t+1 2at. Here the S distance refers to the distance from A to D, and the time is 3 10 seconds.

The initial velocity of v0 is 0. Substitution all to get a = 3 digits reserved for 487. And because vt=v0+at.

The velocity through point d is 146 cm s and is replaced by m s for vt=v0+at and s=v0t+1 2at.

Add the distance of AD The exposure time can be known So the time taken from A to D is known Divide the time T by s to get the average velocity of the whole process Then use the velocity averaging formula V at the end plus v by the beginning of 2 equal average velocity Acceleration can be used V end equals v at the beginning Acceleration into time Having said all this, you still have to calculate it yourself, this problem is very simple. There are also many ways to do this.

All time t=t1+t2=l 3 25+2l 3 50=2l 75

The average speed of the whole course v=l t=75 2=

There are several types of rectilinear motion:

1.Linear motion with uniform velocity, i.e., constant velocity and 0 acceleration

2.Uniform acceleration in a linear motion, i.e., the acceleration is constant and in the same direction as the initial velocity.

3.Uniform deceleration in a straight line, i.e., the acceleration is constant and opposite to the initial velocity.

4. Uniform acceleration linear motion, that is, the acceleration changes uniformly according to the time relationship, and the acceleration may become larger or smaller, but it is in the same direction as the initial velocity.

5. Uniform deceleration linear motion, that is, the acceleration changes uniformly according to the time relationship, and the acceleration may become larger or smaller, but it is reversed from the initial velocity.

6. Variable acceleration linear motion, that is, the acceleration changes according to the time relationship, but it is not uniform, and the acceleration may become larger or smaller, but it is in the same direction as the initial velocity.

7. Variable deceleration linear motion, that is, the acceleration changes according to the time relationship, but it is not uniform, and the acceleration may become larger or smaller, but it is reversed from the initial velocity.

To summarize from the above, accelerating motion, acceleration must be in the same direction as the initial velocity, deceleration motion, acceleration must be reversed to the initial velocity (if the direction of velocity changes in the process, it is best to analyze and calculate the process separately).

The uniform change can be regarded as a one-time equation, for example, the acceleration is y, the time is x, y = ax + b (x is the time elapsed by the motion, y is the time of x, the acceleration of the object, a is the change in acceleration per second, b is the moment 0, the initial value of the acceleration).

According to the formula, the equations are listed, and the problems of linear motion are easy to solve, and the key is to analyze what kind of linear motion each process belongs to.

If you think about the formulas of uniform speed movement, it is better than anything else.

Essence: That is, you are floating in space, not affected by any force, you will have 2 states of movement:

1. Forever still in space; 2. Keep moving in a straight line, and you can only do straight line motion.

Ha ha. You ride a bicycle and walk in a straight line, accelerate, decelerate, and speed at a constant speed.

Just understand the acceleration.

Hello! Because the displacement formula is.

s=v0t+1/2at^2

So vo=6m s

a=-1m/s^2

So it takes 6 seconds to brake!

It is to find what is the value of t when s is taken as the maximum value, 6