High School Exercise Problems, High School Physics Motion Problems

The speed of B at the time of the encounter is 0 Use this to consider the bounty of 0 points who is happy to do.

High School Physics Little White Paper Page 2 Question 8.

Process. The first case: V1 V2, because it is the A ball that is thrown out first, and the initial velocity of the B ball is faster than that of the A ball, so the B ball will definitely meet the A ball in the air, and the height of the upward movement of the A ball is certain, if you want to take the longest time, then the B ball should move at the highest height before the encounter, so when A and B balls meet at the highest point of the A ball movement, the B ball will move for the longest time.

Once the movement situation is known, the calculation process is easy.

The second case: V1 V2, or A ball is thrown first, if the two balls meet in the process of B ball rising, obviously the time is too short (the time of B ball movement is too short), which does not meet the topic. So the two balls will only take a little longer to meet in the process of falling ball B (we all know that the same ball of iron lands at the same time), so if the two balls reach the highest point of their respective motion at the same time, then the two balls will never meet (the relative displacement can be negative under the precursor), so when the A ball is thrown, the B ball is thrown again, and then the two balls reach the highest point of their respective motion at the same time, and then return to fall down at the same time, but they will never meet in the process of falling.

The only one can only meet after reaching the ground, and if there is no ground, then there is no encounter. If this possibility allows, i.e., the two balls will never meet without ground, and the t-value that needs to be calculated is the time difference between the two balls moving to their highest point, i.e., t=(v1-v2) g. The t-value of the first case cannot be represented here, but it is easy to calculate, so it will not be written out here.

Let's add to it...

This question is difficult in the third year of high school, and it also involves some momentum, but I'm too lazy to calculate, I'll talk about a knowledge point, when interacting, the mass and acceleration of 2 objects are inversely proportional, you can set the mass m and km of 2, you know that the acceleration of 2 should be ka and a, and then the isolation method is solved by writing kinematic formulas in the order of development of things given by the question through the procedural method.

It can be known from the speed and time coordinates.

If the car is composed of uniform acceleration and deceleration, it can only run km in 68 seconds, and the time at a constant speed is obviously 28s

The rest of the acceleration and deceleration time is 40s

When acceleration time = deceleration time = 20s

Its acceleration = 100 20 = 5m s

Using the formula s = vmax 2 - vo 2 (2a), we can get s1 = 10000 - 0 2a acceleration phase, s2 = 0 - 10000 -2a deceleration phase, because s1 + s2 = 4800

So 4 * 4800 * a = 20000 can get a value.

The maximum speed on the highway is 120 km/h

The corresponding running distance within 1s is 120km h * 1s;

The brakes of the two cars are the same, so the brake sliding distance of the two cars at the same speed is the same and offset;

Therefore, the 1s running distance is the minimum distance maintained by the two vehicles without touching;

The safety distance is 3 times;

The safety distance is 120km h *1s *3=100m

Set the time of the drop by t

The velocity when the parachute is opened v=gt

42-（gt）2=2ah=-2*2h

h=25t2-4

t = root number 10

Everything else is in the previous steps. 2 is the deceleration time after the value of h t2=(4-gt) (-2).

Free fall time t1, time after opening the umbrella t2, total time t, height h at opening the parachute, free fall 266-h, maximum speed = v

Free fall acceleration = 10, after opening the parachute = -2

2*10*(266-h)=v-square.

V-4 = 2*2*h

The solution yields h = 221 m and v = 30 m sec.

t1 = 30 10 = 3 seconds, t2 = (30-4) 2 = 13 seconds t = 16 seconds.

Free fall time t1, time after opening the umbrella t2, total time t, height h at opening the umbrella

The time of free fall is the time from the time of acceleration descent from a standstill to the time when the parachute is opened, and the gravitational acceleration of this process is g = 10.

After opening the parachute, the athlete performs a uniform deceleration motion with acceleration a=-2.

The velocity when it reached the ground was 4m s, and the entire motion was 266m.

Using the formula for acceleration, A=(st-s0) (tt-t0)4m S=va-2ta

va=0+10t0

266=(4m/s-0)/2*t

t=133s=ta+t0

4=10t0-2ta

12t0=270

to=,ta=

va=4+2*

h=(225-0)/2*

In summary, 1The time for the athlete to fall freely is seconds, 2The athlete opens the umbrella 266-5 = 261 meters above the ground, 3The total time an athlete spends in the air is 133 seconds.

The displacement formula of uniform velocity motion: s=(v2 2-v1 2) (2a) The velocity change is not the same regardless of the same distance or the same displacement.

The relationship between displacement and velocity change can only be explained in terms of velocity variance.

Note: "I don't know the formula at all".

v2=v1+at, so t=(v2-v1) a so:s=(v1+v2) 2*t=(v1+v2) 2*(v2-v1) a=(v2 2-v1 2) (2a).

The answer to 3L is very funny, list a formula and give the answer directly. I didn't see through it at all."

Of course, if you don't understand the expression s=(v2 2-v1 2) (2a), you will find it funny, but if you understand the meaning of s=(v2 2-v1 2) (2a), it is not funny at all. Because, from the formula s=(v2 2-v1 2) (2a), it is obvious:

For a uniform velocity motion, where the denominator 2a is constant, then the distance is proportional to the "variance of velocity" and not to the "difference of velocity".

The relationship between "Variance of Velocity" and "Difference of Velocity" is:

v2 2-v1 2=(v2+v1)(v2-v1) So the same speed change in the same distance is obviously wrong!

From vt 2-vo 2 = 2ax, obtained.

x=(vt^2-vo^2)/2a

And because of the linear motion at a uniform variable speed, a does not change.

i.e. x is proportional to (vt 2-vo 2).

However, if we assume that the velocity changes the same in the same distance, X needs to be proportional to (vt-vo).

Obviously, (vt 2-vo 2) is not equal to (vt-vo), so the proposition does not hold.

Even changing the distance to displacement is not right.

If the velocity change within the same displacement is the same, then it is a uniform linear motion.

However, the velocity changes the same distance and is not necessarily a uniform linear motion.

It should be "same time". The definition is that in variable speed linear motion, the motion with the same velocity change in the same time is called uniform variable speed linear motion. In the linear motion with uniform variable speed, the velocity change of the same displacement or the same distance is not the same, which can be determined by the displacement formula 2as=vt squared-v of the linear motion with uniform variable speed.

squared.

The mistake in this sentence is that it is not the "journey", but the displacement.

How much does its acceleration increase? That doesn't count.

If you increase a n

See figure. <>

s=1/2at² .Boss, you said that it is a uniform acceleration How does its acceleration increase Do you know what is called uniform acceleration linear motion Speechless!!

The results can be obtained by solving the v--t image.

or average velocity solving.

Didn't you give how much acceleration you added?

Solution: If you want to have the shortest time, then there is no uniform process. That is, it accelerates all the time, and decelerates directly when it is added to a certain time.

In this way, set the maximum speed to VM

Then there is: vm 2a1 + vm 2a2=1600 and there is: vm*t 2=1600

From the first formula: vm=64m s

From the second formula: t=50s

That is, the shortest time is 50 seconds, and the maximum speed is 64m sThe easiest way: draw a V-T diagram, calculated according to the area. Hope!

Set the acceleration time to t0 and the deceleration time to t1

s=because the initial velocity is when decelerating, and the average velocity when decelerating to 0 is s=t1=

So. s = solution. t0=40

t1=10, so the shortest time is t=t0+t1=40+10=50 (seconds).

The easiest way to do this is to draw a V-T diagram based on the area.