Tasks on kinematics in which it is necessary to calculate speed time or a way evenly and rectilinearly moving bodies, meet in a school course of algebra and physics. For them find solutions in a size condition which can be balanced among themselves. If in a condition it is required to define time at the known speed, use the following instruction.
It is required to you
- - handle;
- - note paper.
Instruction
1. The simplest case – the movement of one body with the set uniform speed. The distance which the body passed is known. Find time in a way: t = S/v, hour, where S – distance, v – the average speed of a body.
2. The second example - on oncoming traffic of bodies. From Paragraph A the car with a speed of 50 km/h moves to Paragraph B. Towards to it Paragraph B was at the same time left by the moped with a speed of 30 km/h. Distance between Paragraphs A and B of 100 km. It is required to find time through which they will meet.
3. Designate a meeting point by letter K. Let the distance of joint stock company which was passed by the car will be x km. Then the way of the motorcyclist will make the 100th km. Follows from a statement of the problem that time in way at the car and the moped is identical. Work out the equation: x / v = (S-x)/v’, where v, v’ – speeds of the car and the moped. Having substituted data, solve the equation: x = 62.5 km. Now find time: t = 62.5/50 = 1.25 hours or 1 hour 15 minutes.
4. The third example – the same conditions are given, but the car left for 20 minutes after the moped. To define, what is the time in way there will be a car prior to a meeting with the moped.
5. Work out the equation, similarly previous. But in this case moped time in way will be 20 minutes more, than at the car. For equalizing of parts, subtract one third of hour from the right part of expression: x / v = (S-x)/v '-1/3. Find x – 56.25. Calculate time: t = 56.25/50 = 1.125 hours or 1 hour 7 minutes 30 seconds.
6. The fourth example – a task on motion of bodies in one direction. The car and the moped with the same speeds move from A. Izvestno's point that the car left for half an hour later. Through what time will it catch up with the moped?
7. In this case the distance which was passed by vehicles will be identical. Let time in a way of the car will be also hours, then time in a way of the moped will be x+0.5 of hours. At you the equation turned out: vx = v’ (x+0.5). Solve the equation, having substituted value of speed, and find x – 0.75 hours or 45 minutes.
8. The fifth example – the car and the moped with the same speeds move in one direction, but the moped left the point In which is at distance of 10 km from a point And, for half an hour earlier. To calculate through what time after start the car will catch up with the moped.
9. Distance which was passed by the car, 10 km more. Add this difference to a way of the motorcyclist and balance parts of expression: vx = v’ (x+0.5)-10. Having substituted values of speed and having solved it, you receive the answer: t = 1.25 hours or 1 hour 15 minutes.