Questions about acceleration in physics, questions about acceleration in physics

There are two ways to say this:

1. In terms of sections, it is to do a uniform deceleration linear motion first, and after the speed is reduced to 0, then do a uniform acceleration linear motion in the opposite direction.

2. From the perspective of the whole process, it is a uniform and decelerating linear movement.

In linear motion, if it is an accelerated motion, the acceleration is in the same direction as the velocity (this velocity is the velocity at each moment in the corresponding time), and the accelerated motion is of course a linear motion in a single direction.

The object decelerates first and then accelerates.

When it is said that the direction of acceleration of an object is the same as the direction of velocity when it is doing accelerated motion, it means that it is the same as the initial velocity.

Reciprocating motion can also be used, as long as the positive direction is set, if the velocity is negative, it proves that it is already moving backwards.

The car is doing a deceleration motion because the initial velocity is greater than the final velocity.

In an accelerated motion, the direction of the initial velocity is the same as the direction of the end velocity, so the direction of acceleration is the same as both.

Acceleration can be used in vertical upward throwing motion.

First, what is accelerated motion? That is, the absolute value of the velocity is increasing, and in the same way, deceleration means that the absolute value of the velocity is decreasing. And 3 can't jump to -8 right away, so 3 goes down to 0 and then to -8So it's a slowdown first, then an acceleration.

The direction of acceleration when an object is in accelerated motion is the same as the direction of velocity. Velocity refers to the initial velocity.

1.Decelerate first before accelerating.

2.It's the speed at which it's in motion.

3.No. For example, curvilinear motion.

1. Change direction; Displacement. Increase, decrease.

2, The acceleration in uniform linear motion is 0 because there is no change in the magnitude and direction of the velocity.

Thank you for adopting.

Let the initial velocity of an object be v, travel through s1 in time t, and walk through s2 in the next time t, you use the basic formula to list the equation to find s1, and the expression for s2, and then you subtract s1 from s2, you will find s2-s1=at2This is a general result (in uniform acceleration motion).

Then you are doing the lab, the dot timer is dotted according to time t, in a continuous six-stage displacement, set to s1 s6, then s6-s3=s5-s2=s4-s1=3at2

In this way, a=x6+x5+x4-x3-x2-x1 (9t2) is similar to the other segments, but it seems that the number of segments can only be used if they are even, unless you remove a segment and make it an even number.

See the picture? Let me explain, the passenger car starts to move with an initial velocity of 0, and then when it reaches a certain speed (let's set this velocity vt) it starts to decelerate to zero, and the magnitude of the displacement is the size of the area of the triangle, and according to the title it means x and draw a dotted line, and now we can list two equations.

The line on the timeline to the left of the dotted line to the right of the origin is t1 and the stop time is t2So (t1+t2) vt=x

t1=vt÷a t2=vt÷a

The two equations can be obtained by concentrating.

The image is a V-T diagram in a triangle, the slopes of the two waists are two accelerations, the area of the triangle is the displacement x, and then the area is represented by two displacement formulas.

The first thing you have to know is that the first thing the car does is to move in a straight line with uniform acceleration, and the traction force is constant during this time, and when p increases to 60 kWh, p does not change, and f starts to decrease, and when it is so small that it is equal to the drag, it starts to move in a straight line at a uniform speed.

vmax=p fmax=60000 5000m s=12m s, i.e. f-5000=5000*

Find f=7500n

v=p/f=60000/7500m/s=8m/st=v/a=8/

1.When f=f friction, p=fv, v=12m s

f(resultant) = 2500, p = fv (this f = 2500 + 5000 = 7500), v = 8, v = at, t = 16s