A car moves in a straight line at a constant speed of 15 meters per second

Assuming that the time required for the police to chase is t, then the distance of the car from place A when catching up with the car is .

s=（ （1）

At the same time, the distance traveled by the motorcycle is also s, s=2*t 2 2 (2) is solved by (1) (2) formula.

t 16, i.e. you can catch up with the car after about 16 seconds.

Set the time to t15*

t=---haha, many people don't understand!! Landlord, do it yourself!

Vehicle displacement: s1=vt

Displacement of police car: s2 = 1 2at 2

Start spacing: l=vt0

s1+l=s2

vt+vt0=1/2at^2

15t+15*

t solution t1 = (15 + 291 under the root) 2, t2 = (15 - 291 under the root) 2t2 = (15 - 291 under the root) 2 is not in line with the topic, discard.

So t=(15+291 under the root) 2

Set t seconds to catch up.

The distance traveled by the car is:

Frequently moved journeys for:

After reciprocity. t is approximately equal to 16 seconds.

Let the time t and list the equation:

Solution t=seconds.

The distance traveled by the car s1 = 15m s*

The distance traveled by the police car is s2 = 1 2at*t=1 2*2 meters per square second * t squared.

S1=S2 fractions and units are not easy to play, I don't know if I can understand this.

Set x seconds later the police catch up with the car.

square - 15*x=

Just solve the x again.

Set to catch up with the car after X seconds.

15 * x + = 1 2 * 2 * x 2 gives x= ......

Let's figure it out for ourselves.

Let the time be t, then there is.

15*（t+

Just calculate t.

Summary. The distance in two minutes is kilometers. Equation 5 2 60 1000 = km.

If a car moves in a straight line at a constant speed of 5m per second, how many kilometers + how many kilometers will it pass in two minutes.

The distance in two minutes is kilometers. Equation 5 2 60 1000 = km.

Because one minute is equal to 60 seconds, the second branch guesses the clock is 2 60 = 120 seconds, because every second of the fierce macro type drives 5 meters, so 5 120 = 600 meters. 1 km = 1000 meters, so 600 1000 = km.

The combined formula is 5 2 60 1000

So how many seconds does it take for the car to travel 3km?

Time required 600 seconds, equation 3 1000 5 = 600 time = distance The speed of 3 km is 3000 meters.

In the 200m competition of the middle school students' sports posture and fiber luck book, Xiaohua ran a time of 25 seconds, so the average speed of his running is more than 100 meters per second and how many kilometers per hour?

His average speed is 8 meters per second. Eq. 200 25 = 8 kilometers per hour Eq. 8

Summary. A car starts from a standstill, does a uniform acceleration with zero initial velocity, and moves in a straight line after the speed is increased to 25 meters per second.

Hello, what is the specific question?

90m/s=25m/s

A= is derived from at=25

The velocity at the 15th is v=0*15+

The displacement of 15s is: s1 = 1 2*at1 2 = 1 2*, and the displacement of 25m45s is: s2=vt2=

The displacement in 1 minute from the start is: s=s1+s2=the average velocity in this time is: v'=s/t=

90km h=25m s is obtained from at=25 a= velocity at the 15th s is v=0*15+ displacement at the 15th s is s1=0*15+1 2* So the displacement of the uniform linear motion of the car in the last 45 seconds is s2=45*The total displacement in minutes is s=s1+s2=the average velocity is v'=s/60=

After 10 seconds, the velocity is 25m s, which means that the acceleration a 25 10= After another 5 seconds, the velocity of the acceleration that is still accelerating in these 5 seconds is, then the speed of doing the uniform linear motion is.

lz, the first burning stool is to ask when to destroy the farthest away from the Zen, right?

Motorbike departing t=0

It is the farthest away from the rest of the journey.

First question. t=10/2=5s

The distance between vt + 11-2 1at 2 = 36m

vt+11=2/1at^2

Solution. t=11

VT+11=121m from place A

The distance of motion during this time is s=1 2at 2 and then the square of the final velocity minus the square of the initial velocity is equal to the two keys and the multiplier acceleration times s(v1 2=2as+v2 2).

Note: v1 = the speed at the end of the second of the three worms.

v2 = initial velocity.

Find the average velocity 120 10=12 first

The first five seconds of travel is equal to 12*5=60 meters.

3km=3000m 5min=300s so the first empty is 3000 300=10

Since it is a constant motion, the speed has always been the same, 10m s should be 10*60*60, 1000=36km h

If a car in a straight line with a uniform velocity travels 3 km in 5 minutes, the speed of the car is 3000 5 60 m s, and the speed of the car at the end of the 5s is 3000 5 60 km h

Children, such a simple oral arithmetic problem, I suggest you still do it in a down-to-earth way, think about it, since you can go online, you can type so many words, it is not a problem to do this question, it may be more difficult in the future, after going to college, and even after work, you will encounter a lot of problems, not everywhere there are ready-made answers, only on your own, although you and I don't know, but I sincerely advise you to really realize this, come on.

From 3000 (5*60), the speed is 10m s

Since it is a constant velocity, the speed remains the same at the end of the fifth second, and the speed is still 36 km/h