A physics problem. A car has a speed of 90 km h before braking, and the acceleration obtained by bra

It's very simple, just set the formula s=.

1) Substituting numbers into it and solving equations.

2) Find the time at rest t1=v a=9s

Then x=(v*).

1) The deceleration process is a uniform variable speed motion with initial velocity v0=90km h=25m s acceleration a=-10m s 2.

s=v0t+ solution to obtain t=2s or 3s, where, t=3s is the time of reciprocating motion, and the actual forklift process does not have this process, so t=2s

2) Study the reverse process of braking, and do uniform speed movement with a=-10m s 2 without initial velocity. x=

1. First calculate the time taken to stop, and then calculate the displacement used to stop. Compared to 30 meters, 30 meters is the time taken to stop. Otherwise, solve the 2nd order equation and bring it into solution. 2. With reverse thinking, re-stationary begins to seek displacement, acceleration, and time, is it still not possible?

There are example problems in advanced mathematics, which is a typical definite integral application problem. It is recommended to turn the book, "Advanced Mathematics" Volume I, page 292.

First 90km h = 25m s

1）30=25*t-1/2*10t^2

The solution is t=2s (obtained by v=v0+at, and the car stops in seconds when braking) 2) the displacement in the first second before stationary can be considered with a uniform acceleration linear motion in the opposite direction 1 2*10*1 2=5m

1.According to s=vt+1 2at 2, 90km/h = 25m s30=25t-1 2*10*t 2 t=2 or 3 But since it is 3 seconds, the speed of the car is already 0, which is not true, so t = 2 seconds.

2.By the initial velocity path and acceleration of the car, according to the formula V end = V beginning + AT end;

You can find the second when the car stops.

The second before the car comes to rest is the second. When the car speed is 15m s according to s=vt+1 2at 2

s=15*1-1/2*10*1^2=10m

1) Let the velocity after displacement of 30 meters be v

90*90-v*v=2*10*30；Solve v;

t=(90-v)/10

2) x=(10*10)/(2*10);Solve x

72km/h=20m/s

Braking time t=v0 a=20 4=5sThat is, the car stopped in 5s.

The displacement of the brake 10s is actually the braking distance split, s=v0 2 2a=400 8=50m

s=, i.e.: 30=20t-2t2, and t=5-sqrt(10)s

Braking can be seen as an "inversion" of the uniform acceleration motion from the stop positionSo the distance of glide before stationary is 1s'=,2,Rainbow fluttering report.

What does t=5-sqrt(10)s mean oao 5 minus the root number 10, which is approximately equal to, the speed of a car before braking is 72km h, and the acceleration obtained by braking is 4m s2, find:

1) The distance of the front hand within 10s after the car starts braking is x0?

2) The time experienced by Bi Xian from the start of braking to the displacement of the car is 30m t (3) The distance x of the car coasting within 1s before the car is stationary'

Question (1) 1, vehicle speed v=90 meters and seconds).

2. First of all, in addition to special tires, the maximum deceleration of the car cannot reach 10m s, except for high-grade special tires.

3. The time when the car stops, 25 10 = seconds).

4. So in 10 seconds the glide distance has stopped in seconds. s=25*25 2 10=m).

Question The average velocity of the meter before stopping, [25+(25-10*t)] 2=25-5t

2. From 30=(25-5t)*t, it is obtained, 5t*t-25*t+30=0; t²-5t+6=0;(t-2)*(t-3)=0, get t=2 or t=3, due to the previous analysis, the car has stopped in seconds, saying that t=3 is not true, it can only be established in t=2 seconds.

Question (3) 1, the speed before parking is 10 meters per second.

2. The distance of the car sliding within 1s before stationary, s=10*10 2 10=5 meters.

(1) (2*10) = m.

2) Inverse operation: (90 root number = 2 seconds.)

3) From the end point of the speed of 0 is calculated from 1 2 * 10 * 1 2 = 5 meters.

The initial velocity is 25m sI count it after the second stops.

diy。Won't ask again.

Before braking, the speed is converted to v0=25m s acceleration a=-10m s

1): From v=v0+a*t, it can be calculated that the time taken to brake to speed is zero 0=25+(-10)*t t=

by v2 -v1 =2ax 0 -25 =2*(-10)*x x=

2): By v2 -v1 = 2ax when x = 30, v2 -25 = 2 * (-10) * 30 v2 = 5m s

When v=v0+a*t v=5m s, t=2s(3): from the first question, we can know the common need to stop, and the distance traveled first is obtained by x=v0*t+ x=

So the distance for the car to coast within 1 second before stationary is x=

Please calculate according to the formula. s=vt+1 2(a 2)t can be obtained by substituting the data.

1) Coasting 125m, because the car is actually only used when stopping, so the time taken to drive 125m (2) is 2s, and there are two answers, but the car stops only with 2s

3) The gliding distance is 5m, and the speed at the previous time is 10m s, so the sliding distance is 5m for 1s after the pie is scattered

90km/h=25m/s

a) By the theme of the car brakes seconds to stop.

So the displacement in 10 seconds is.

25*2) 30=25*t-1 splitting 2*10t 2 solution t=2 3) before stationary Yu Yuan attacked one second Qing brother shift 1 2*10*1 2=5m s

(1) Test question analysis: (1) Judge the time experienced by the car braking v090km h=25m s

by 0=v0+at

and a=-10m s2

Get: t=-v0

a=25/10=<10s

The car brakes and comes to a stop, so the displacement of the car within 10s is only the displacement within.

Solution 1: Use the displacement formula to solve.

= Solution 2: According to the <>

Got: <

2) According to <>

Got: <

Solution: t1=2s, t2=3st2

It is a car by T1

After continuing to reach the farthest point, and then the reverse acceleration motion re-reaches the displacement is the time elapsed when it is 30m, since the car brakes are one-way motion, it is clear that t2

If it doesn't fit the topic, it must be discarded.

3) Solution 1: The process of decelerating the car to zero velocity is regarded as the process of uniform acceleration motion with zero initial velocity, and the car is found to be 10m s2

The acceleration is displaced by 1s, i.e., <>

5m solution 2: the end of 1s before stationary is the end after the start of deceleration, and the velocity at the end is v=v

at=25-10×

So: <>

5M Comments: When analyzing the linear motion of uniform variable speed, there are many methods because there are many rules, and two different methods can be used to compare the calculation results.

v=v0+at

The final velocity v=0 then 0=v0+at

So t=-vo a=-8 (-6)=4 3(s) displacement x=v0t+

First, calculate the stop time: 0 8 6t to get: t 4 3 4 seconds. It has been stopped 4 seconds ago. Displacement: s 8 4 3 1 2 6 4 3 square 16 3 m. Thank you.

16/3 m

vt^2-v0^2=2as

It's easy with this formula.

When encountering the type of brake, we must first calculate how many seconds to stop v=v0-at, so t is equal to 4 3 seconds, so the car stops in 4/3 of a second, and we can get x=49 9 by using the displacement formula x=v0t+1 2at2

The time it takes to reduce the speed to 0 t=(8-0) 6=4 3 seconds After braking for 4 seconds, there is actually (4-4 3)s that the time is stationary. Displacement s=8*(4 3)+