A and B are two kilometers apart, and the car is traveling at a speed of 24 per hour

Solution: 180 3 = 60km hours.

A: The speed of the car is 60km hour.

(1) t=s/v=100/v

2) t=100/(v+5) t1=100/v-100/(v+5)

3) t1=100 50=2 hours t2=100 (50+5)=hours t3=hours.

Time is equal to the distance divided by the speed, and the higher the speed, the less time.

1, the car from A to B needs to travel time t1 = 100 v2, if the speed increases by 5 kilometers hour, then the required time t2 = 100 (v + 5) t1 - t2 = 100 v - 100 (v + 5) 3, v = 50, then t1 = 100 50 = 2 (hour) t2 = 100 55 = 20 11 (hour).

t1-t2=2 - 20 11=2 11 (hours), that is, the increase in speed is 2 11 hours earlier than the original.

240 60 = 4 (hours).

7+4=11 (hour).

A: The car arrived at place B at 11:00

Solution: s=60t (0 t 5).

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(1) According to the title, the car should be driven at a constant speed, and the speed is v km/h, then the whole journey must take time t = s v hours.

y = (a + bv )t = (a + bv )s v = s(a v + bv) yuan.

2) y = s(a/v + bv) ≥2s√[a/v)(bv)] = 2s√(ab)

A v = BV (v = (AB)), the whole journey cost is the most economical.

Solution (1):

h (h t) hours later, the distance between the two cars is: (s-ah-bh ) km (2):

After h hours, the distance between car A and place B is: (s-ah) kilometers (3):

At the time of the encounter, car A traveled the following distance: at kilometers; The distance traveled by car B is: BT kilometers (4):

At the time of the encounter, car A travels more than car B for (at-bt) kilometers.

420 (60+80) 420 140 3 hours A 3 hours meeting 420 (60+80)=3h 20, A and B cars travel in opposite directions from 325 kilometers away at the same time, car A travels 52 kilometers per hour, car B,