The car travels on a straight road, and in the process of its speed increasing to V with zero。。。。。

Analysis: The amount of work done by the force on the object is determined by the force and the displacement, and the calculation of the displacement often needs to be solved by kinematic formulas.

Analysis: Since the traction and resistance of the automobile engine are unchanged, the movement of the automobile is a linear motion with uniform speed. Let the acceleration of the car be a, the displacement from zero to v is s1, and the displacement from v to 2v is s2, then there is:

2as1 v, 2as2 (2v) v so there is s2 3s1

then w1 fs1 ,w2 fs2 f·3s1 3w1

Answer: B

The answer is: b let the mass of the car be m, the traction force is f, and the drag force of the car is f, since the traction force and resistance of the car are unchanged, then according to Newton's second law:

f fit ma f f

It can be seen that the automobile is moving in a linear motion with uniform acceleration, and according to the law of conservation of energy, the work w1 done by the automobile engine when the automobile speed is from zero to v is: w=1 2mv fs1

and v 2as1

2v）²－v²＝2as2

Get s2 3s1

When the speed of the car rises from V to 2V, it is obtained according to the theorem of the law of conservation of energy

w2＝1／2m（2v）²－1／2mv²＋fs2＝3／2mv²＋3fs1＝3w1

Therefore, choose B

Let the total distance be s, and the first 2 and which 3 distance times will be.

t1=(2 3s) v1, the remaining time of 1 3 travel.

t2=（1/3s）/v2

The whole journey of the car.

Average. v=s (t1+t2), substituting the data to get 28km h=s 2 3s) brigade v1

1 3s) 20km lifting town yard h;

So v1 = 35km h

Let the total distance be s, and the distance traveled by 2 3 at speed v1, then the time of the first 2 3 journeys of the hand-pinned sail is t1=(2 3s) v1, and the time of the remaining 1 3 distances is t2=(1 3s) v2

The average velocity of the whole process is v=s (t1+t2), and the data is substituted to fight this.

28km/h=s/[=2/3s）/v1+（1/3s）/20km/h]；Solution.

v1=35km/h

It seems that only the quarrel is junior high school physics, and let's see if it's a round belt?

The average speed of the first leg is (0+v) 2=

In the second section, the average speed is (v+2v) 2=

And w (work done) = f (force) * s (distance).

The distance is equal to the time multiplied by the velocity, the force is equal, and the acceleration time is equal, so it depends on the velocity, so the above answer holds.

Solution: (1)x=v0*t+

30m=t=10s.

2) vt=v0+at

0=t=The car stopped at the second.

vt²-v0²=2axx=。

It just so happened that we had just taken this question

1) Use x=v0t+at square 2 to get the time and so on in Congyu 10s2) because when the car stops, the time is seconds. So the time can't reach 20. Therefore, it is used to calculate.

Use x=(v0+v)t 2....Get the distance for.

。Ask for seepage celery rolling.

Let the total distance be s, the time of the first 2 positive 3 journeys is t1=(2 3s) v1, and the time of the remaining 1 3 journeys is t2=(1 3s) v2

The average speed of the car is v=s (t1+t2), and the data brigade is prepared for 28km h=s (2 3s) v1 + 1 3s) 20km h;

So v1 = 35km h

If the total distance is 1, then the time to complete the journey is.

t = 2 3) v1 + 1 3) 20 average velocity = 1 t = 28, t = 1 281 28 = 2 3) v1 + 1 3) 20 can be solved v1

Solution: (1)x=v0*t+

30m=t=10s.

2) vt=v0+at

0=t=The car stopped at the second.

vt²-v0²=2axx=。

When the speed of the car reaches 10m s.

Time t=10

Displacement x=1

Constant speed driving displacement x=10x15=150m

Braking time 10 2 = 5s

Displacement in 5s x=10x5-1 2x2x25=25m, the car advanced 100+150+25=275m

Advance 20 + 15 + 5 = 40s

Solution: t1=10 acceleration time) t2=10 2=5s (deceleration time) teven=15s

s1=1\s2=10x15=150m

s3 = 10x5-1 2x2x5x5 = 25ms total = 275m

t total = 40s

D choose DThe car does the acceleration motion of variable traction force under the action of constant power, so the work done by the engine is the work done by variable force, and W pt can be obtained according to p w t, and p f v fvm

So w fvm

t；Conservation according to energy: w <>

mv<>

mv<>

fs, so w <>

mv<>

fs－<>

mv<>

Figure 4-1-6