Instruction
1
Let us examine how to solve common tasks.
The solution of problems on velocity we need to know a few formulas and be able to correctly write the equation.
Formula for the solution :
S=V*t formula is the way;
V=S/t formula speed;
t =S/V time formula, where S is distance, V - velocity, t - time.
For example, let us consider how to solve tasks of this type.
Condition: the Truck on the way from city a to city B spent 1.5 hours. The second truck has spent 1.2 hours. The speed of the second car more than 15 km/h than the speed of the first. To find the distance between two cities.
Solution: For convenience, use the following table. In it, specify what is known by the condition:
1 auto 2 auto
S X X
V X/1.5 X/1,2
t 1,5 1,2
For X take what you need to find, i.e. the distance. When setting up the equation to be careful, note that all values were in the same dimension (time - in hours, speed in km/h). By the speed of the 2nd car for more speeds 1 to 15 km/h, i.e., V1 - V2=15. Knowing this, we form and solve the equation:
X/1,2 X/1,5=15
1.5 X - 1.2 X - 27=0
0.3 X=27
X=90(km) - distance between cities.
Answer: the Distance between the towns is 90 km.
The solution of problems on velocity we need to know a few formulas and be able to correctly write the equation.
Formula for the solution :
S=V*t formula is the way;
V=S/t formula speed;
t =S/V time formula, where S is distance, V - velocity, t - time.
For example, let us consider how to solve tasks of this type.
Condition: the Truck on the way from city a to city B spent 1.5 hours. The second truck has spent 1.2 hours. The speed of the second car more than 15 km/h than the speed of the first. To find the distance between two cities.
Solution: For convenience, use the following table. In it, specify what is known by the condition:
1 auto 2 auto
S X X
V X/1.5 X/1,2
t 1,5 1,2
For X take what you need to find, i.e. the distance. When setting up the equation to be careful, note that all values were in the same dimension (time - in hours, speed in km/h). By the speed of the 2nd car for more speeds 1 to 15 km/h, i.e., V1 - V2=15. Knowing this, we form and solve the equation:
X/1,2 X/1,5=15
1.5 X - 1.2 X - 27=0
0.3 X=27
X=90(km) - distance between cities.
Answer: the Distance between the towns is 90 km.
2
In solving problems on motion on the water " you must know that there are several types of velocities: the velocity (V) speed current (Vпо tech.), speed against the current (V [PR]. tech.), flow rate (Vтеч.).
Remember the following formula:
Vпо tech=VC+Vтеч.
V [PR]. tech.=VC-Vтеч.
V [PR]. tech=tech Vпо. - 2Vтеч.
Vпо tech.=V [PR]. tech+2Vтеч.
VC=(Vпо tech.+V [PR] tech.)/2 or VC=Vпо tech.+Vтеч.
Vтеч.=(Vпо tech. - V [PR]. tech)/2
For example, let us consider how to solve them.
Condition: boat Speed the flow of 21.8 km/h and upstream 17.2 km/h to Find a private speed boat and speed of river flow.
Solution: According to the formula: VC=(Vпо tech.+V [PR] tech.)/2 and Vтеч.=(Vпо tech. - V [PR]. tech)/2, we find:
Vтеч = (21,8 - 17,2)/2=4,6\2=2,3 (km/h)
VC = V [PR] tech.+Vтеч=17,2+2,3=19,5 (km/h)
Answer: Vc=19,5 (km/h), Vтеч=2,3 (km/h).
Remember the following formula:
Vпо tech=VC+Vтеч.
V [PR]. tech.=VC-Vтеч.
V [PR]. tech=tech Vпо. - 2Vтеч.
Vпо tech.=V [PR]. tech+2Vтеч.
VC=(Vпо tech.+V [PR] tech.)/2 or VC=Vпо tech.+Vтеч.
Vтеч.=(Vпо tech. - V [PR]. tech)/2
For example, let us consider how to solve them.
Condition: boat Speed the flow of 21.8 km/h and upstream 17.2 km/h to Find a private speed boat and speed of river flow.
Solution: According to the formula: VC=(Vпо tech.+V [PR] tech.)/2 and Vтеч.=(Vпо tech. - V [PR]. tech)/2, we find:
Vтеч = (21,8 - 17,2)/2=4,6\2=2,3 (km/h)
VC = V [PR] tech.+Vтеч=17,2+2,3=19,5 (km/h)
Answer: Vc=19,5 (km/h), Vтеч=2,3 (km/h).
3
Objectives to compare quantities
Condition: Lot of 9 bricks 20 kg more than the weight of one brick. Find the mass of one brick.
Solution: let X (K), then the mass of 9 bricks 9X (kg). From the condition it follows that:
9X - X=20
8x=20
X=2,5
Answer: the Weight of one brick is 2.5 kg.
Condition: Lot of 9 bricks 20 kg more than the weight of one brick. Find the mass of one brick.
Solution: let X (K), then the mass of 9 bricks 9X (kg). From the condition it follows that:
9X - X=20
8x=20
X=2,5
Answer: the Weight of one brick is 2.5 kg.
4
Task on fractions. The main rule when dealing with such this type of tasks: to find a fraction of the number, this number multiplied by this fraction.
Condition: the Camper was on the road 3 days. The first day was it? all the way, in the second the remaining 5/9 of the way, and on the third day - the last 16 km to Find the entire path of the tourist.
Solution: Let the entire path of the tourist is equal to X (km). That first day was it? x (km) second day - 5/9(x -?) = 5/9*3/4x = 5/12x. Since the third day it was 16 km, then:
1/4x+5/12x+16=x
1/4x+5/12x-x= - 16
- 1/3x=-16
X=- 16:(-1/3)
X=48
Answer: All the way a tourist is 48 km away.
Condition: the Camper was on the road 3 days. The first day was it? all the way, in the second the remaining 5/9 of the way, and on the third day - the last 16 km to Find the entire path of the tourist.
Solution: Let the entire path of the tourist is equal to X (km). That first day was it? x (km) second day - 5/9(x -?) = 5/9*3/4x = 5/12x. Since the third day it was 16 km, then:
1/4x+5/12x+16=x
1/4x+5/12x-x= - 16
- 1/3x=-16
X=- 16:(-1/3)
X=48
Answer: All the way a tourist is 48 km away.