Senior 1 Physics, please explain how to arrive at the answer.

Solution: The law of motion of A: vt= 40-10t; s=40t-5t^2;Note that the exercise time is 0 seconds.

The law of motion of B: v't= 20+(20 3)(t-2); s'= -20(t-2)+(10/3)(t-2)^2;Note that the exercise time is 2 seconds.

Two elements, one time variable, the degree of correlation is a mixture of multiplication (including power) and addition.

Therefore, after image comparison, there is velocity intersection below the horizontal axis, so the two particles in the problem have the same time point in the direction of motion;

When the two velocities are equal, they are calculated as seconds in time; Set the distance to d is-s'i i[40t-5t 2] [20(t-2)+(10 3)(t-2) 2]i=i-25t 2+220t-160i 3, it is obvious that this is the minimum value problem of the quadratic function in junior high school, that is, at t= b (2a), there is, the solution, the solvable t = seconds, so the farthest distance between the two particles in the problem is seconds, and the speed is equal. Choose B to finish.

For option a: the above the timeline is in the positive direction, because the velocity is a vector, so there is a direction, and because it is on a straight line, its direction is either the same as the positive direction, or the opposite direction, the same is positive, and the opposite is negative, as can be seen from the figure, both have a negative segment, so there are times when the direction is the same. For option b:

There are three situations: when the two move in opposite directions, the distance between the two is getting more and more ( ) When the two move against each other, the distance between the two is getting more and more ( ) When the two move in the same direction, it belongs to the pursuit problem, the speed of the pursued is large, the speed of the pursuer is small, and the distance between the two is more and more ( ) On the contrary, the speed of the pursued is small, and the speed of the pursuer is large, then the distance between the two is more and more ( ) Understand this is easy to judge, because it is a multiple-choice question, so this method is more convenient.

It is better to teach a man to fish than to teach him to fish. Landlord, let me tell you to think about it and do it yourself.

1) When the ball swings to the bottom d, it is subjected to two forces: the tension of the line and its own gravity, because it is only the bottom point of the line, we make the point d the critical point of the force strength of the line. The following is the force when analyzing the line D point:

The line is subjected to ball gravity; Because the line also gives a centripetal force to the ball, it is subjected to the reaction force of the ball f=(v 2 r)*m....Because the line is just broken at point d, the ultimate strength of the force = + the speed of the landlord's ball to point d is calculated by the conservation of energy (conservation of mechanical energy).

2) If it just does not derail, we will analyze the force state of the ball at the top point, the gravity of the ball completely provides the centripetal force and is not affected by the orbit. So we get (v 2 r)*m=mg. where we find v and thus get the energy e (kinetic energy and potential energy) at the top.

The next only need is to let the energy of the ball at point d - the energy expended on the flat orbit = e. The coefficient of friction determines the amount of energy consumed, and the coefficient of friction is found accordingly.

Solution: Let the velocity of the ball be v when it passes through point d, the velocity when it passes through the highest point of the circular orbit is u, the starting height of the ball is easy to know is 1m, then mgh=mv 2) 2, and the centripetal force is f=mv 2) l, the simultaneous equation knows f=5n, if the ball passes through the highest point and does not fall off, then its centripetal force is all provided by gravity, mgh- mgs=2mgr is known by the kinetic energy theorem, and the equation is known as =0,2, so the friction factor is the maximum.

(1) The total work done by the ball with the kinetic energy theorem is (L-Cosl) mg=1 2mvd square to find vd with the centripetal force formula f max = m to find vd square l (2) smooth arc remember that there is no friction, so at the highest point mg = mv highest square r to find v maximum According to the kinetic energy theorem, the work done by the ball is mg2r = 1 2mva square (this is the limit method), that is, the velocity of the ball at point A when it is just past the highest point is va, so the velocity of the ball at point A is greater than or equal to va According to the kinetic energy theorem, if the friction coefficient of the ball is u on the horizontal plane, then mgus=1 2mva square-1 2mvd square (vd has just been found), to u, according to the velocity of va, the magnitude relation can be until the magnitude relation of va can be found (algebra is enough).

It's still a matter of centripetal force.

When sliding to the lowest point, n-g=mv 2 r, n=mv 2 r+g=mv 2 r+mg

f=un=u（mv^2/r+mg）

ta=f′+mgsinθ ？It is because the object is moving upwards and is subject to friction in the opposite direction given to it by the inclined plane. The object moves upward at a uniform speed: it means that the object is balanced by force in the direction of the inclined plane, that is, the net force is zero, and along the inclined plane: the object is acted on by three forces, 1 is the downward frictional force, the second is the downward component of gravity, and the third is the upward tensile force t.

Therefore, Ta (force up the inclined plane) = f + mgsin (force down the inclined plane) to make A slide up the inclined plane at a uniform speed.

An object of mass m can slide at a constant speed when placed on an inclined plane with an inclination angle of =37°. Indicates that there is sliding friction.

After changing the direction of motion, there is still sliding friction, but the direction is changed, and it is the same as the direction of gravity, so f is added

There is also friction, mgsin is only the component of gravity along the inclined plane, opposite to the direction of the pulling force, there is also a frictional force down the inclined plane.

A d I can't pass it**, I can only use text!

Let's start with the force analysis:

The boat is subjected to a total of 4 forces, where the vertical direction is gravity g downward, the buoyancy force f floats upward, the resistance of the water f resistance is backward, the rope pull force f is at an angle with the horizontal plane , we decompose f, the horizontal direction is f1, and the vertical direction is f2.

Since the ship is moving at a constant speed, then f1=f resistance in the horizontal direction.

And f=f1 cos gets bigger as the ship gets closer to the shore, then f gets bigger, and you can choose a.

In the vertical direction, g=f float + f2 where f2=f1 tg becomes larger as the ship approaches the shore, then f1 becomes larger; In the vertical direction, the force is balanced, g does not change, and f1 becomes larger, then the buoyancy of the ship becomes smaller, so D is selected.

1. The tensile force of the rope (t1 t2) becomes smaller, because the tensile force t = gravity g sina, a (0, 2), a then sina, t

Answer: D<>

A look** Thank you The circle is the boat The horizontal line above is the pulley Thank you I hope you can understand.

AD bar, probably. I do it by breaking it down vigorously. The resultant force is the direction of the boat's motion, the pulling force of the rope is one of the components, and then another force is drawn using the parallelogram rule. An angle is determined, and the change of sin and COS is determined through the change of the angle.

It's better to draw a velocity --- time image so that it's easier to understand.

1) Find the acceleration of the acceleration phase according to the velocity at the end of 2 seconds and the end of 3 seconds, a=v--v0 t=(9--6) 1=3m s2

Based on the maximum speed of 12m s, and, the car must be in the deceleration phase, so the acceleration in the deceleration phase is A'=(v--v0)/t=(3--9)/1=-6m/s2.

2) According to the end velocity of the uniform deceleration phase is zero, the time from the time to the stop is studied. The initial velocity is v0=3m sIf the final velocity is v=0 and the acceleration is a=-6m s2, then this time is t=(v-v0) a=

The total time can be found as t total =

The time when the bullet travels a diameter distance, the paper tube turns at an angle

The result is kept to three decimal places: , instantaneous velocity is distance divided by time, and acceleration is the difference between velocity divided by time.

by s1=v0t1+at1 2 2

s2= v0t2+at2^2/2

Find v0 and a

VA=V0+ATD is then calculated