A few math problems are to chase the problem, ask for a math problem, chase the problem

According to the title, it took a total of 3000 200 = 15 minutes.

Parking for two minutes, that is, delaying 200 * 2 = 400m per minute of more than 100 meters, then the time to arrive on time after meeting acquaintances is 400 100 = 4 minutes.

That is, he arrived on time four minutes after meeting an acquaintance, and the distance was: (200+100)*4=1200m

The distance is: (36+.)

1 From the meaning of the title, it took a total of 3000 200 = 15 minutes.

Stop for two minutes, that is, delay 200 * 2 = 400m, every minute more than 100 meters, then meet the acquaintance after the time to arrive on time is 400 100 = 4 minutes, that is, he meets the acquaintance four minutes after the arrival on time, the distance is: (200 + 100) * 4 = 1200m

I'm only in my first year of junior high school. I also have a headache for topics like yours. Hehe! But I've found a confidant like me.

Question: The brothers set off from home to school, the younger brother walked 50 meters per minute, the elder brother rode a bicycle for 200 meters per minute, and after the younger brother walked for 15 minutes, the elder brother was able to catch up with the younger brother after a few minutes away from home.

Analysis: This is a simple catch-up problem, and the basic relationship is: catch-up distance = speed difference x catch-up time.

According to the title, the distance that the younger brother walked in the first 12 minutes is the distance between the two in the process of chasing, that is, 50 12 = 600 (meters), and the speed difference is 200-50 = 150 (meters), so the chasing time is: 600 150 = 4 (minutes).

Answer: 50 12 (200-50) = 4 (minutes).

The formula for the catch-up problem: Catch-up distance = (fast - slow) x catch-up time Catch-up time = catch-up distance (fast - slow) catch-up speed = catch-up distance Catch-up time.

Very good, but there is no answer to the back.

French rain basic v pay oh at all.

Distance difference = speed difference time.

What is the correct formula for calculating the catch-up problem?

When the first x meters of A are missing, B starts from the starting point, and the two arrive at the end at the same time, and the time is the number of leases t can be the equation. 500=1×t

500=2×(t-x/1)

Solve the binary equation together to get x = 250 meters.

There's generally no formula to follow, but there are rules that can be found by looking at the time interval or the difference in distance or the speed, and the methods are different.

Generally the time is equal. The equation is then listed according to the distance.

Chase and problem Chase and distance Speed difference Chase time Chase time Chase and distance Speed difference Speed difference Chase and distance Chase and time.

This is the principle, and it is very simple to figure it out according to this!

You don't know how to take pictures, so there's no way to do the math for you.

There are only two cases of follow-up problems, and there are 2 variations in each case. As long as you have a firm grasp of these two situations, you can easily solve the problem of chasing.

1. Introduction to the question type.

The catch-up problem is a typical application problem of the common examination of the travel problem, which is a problem that studies the "same direction motion", and the catch-up problem reflects the relationship between the distance, speed and time traveled by two or more quantities. The core is the speed difference.

2. Core knowledge.

Catch-up time = distance difference, speed difference;

Distance difference = catch-up time, speed difference;

Speed difference = distance difference Pursuit time.

Xiao Hong and Xiao Ming's homes are 300 meters apart, and the two set off from home to school at the same time, Xiao Ming is behind Xiao Hong, and Xiao Ming is every minute.

Zhong walks 160 meters, Xiaohong walks 100 meters per minute, ask Xiao Ming how many minutes to catch up with Xiaohong?

Catch-up time = distance difference Speed difference = 300 (160-100) = 5 minutes.

3. Detailed explanation of the use of core knowledge.

When the catch up problem occurs in a straight line: distance difference = chaser's distance - chaser's distance = speed difference and chasing time;

When it occurs on a circular distance: fast distance - slow distance = circumference of the curve;

It is a question of how two objects can meet when they start from different or the same place. Starting from the same place, the conditions can be basically converted to: the distance of the motion of two objects is the same or a certain multiple; Starting from different places, the conditions are changed to:

The question of how much is the sum of the distances traveled by two objects. In fact, this kind of problem is not complicated, and most teachers make it too complicated. Just connect more with reality, think about how they met, and what is the mathematical relationship between the distances they travel.

In motor racing, we often see the fierce scene of several cars chasing, can the car behind catch up with the car in front? This is called the catch-up problem in mathematics, and it refers to two objects moving in the same direction, with the faster one behind the slow one, and after a certain period of time, the faster one catches up with the slow one.

It's a kind of problem where two people have a difference in velocity and displacement at the same time, and they determine the location of the meeting by discussing factors such as time.

Thank you for your help, I have a pass in the exam tomorrow.

Solution: Derived from the question.

A hound's 12-step run is equivalent to a rabbit's 32-step run, and a dog's 12-step run is equivalent to a rabbit's 27-step run.

So for every 27 steps the rabbit runs, the hound catches up with 5 steps (rabbit steps).

The hound needs to run to catch up with 80 steps (rabbit step).

27 (80 5) 80] 8 3 192 (step) Answer: The hound must run at least 192 steps to catch up with the hare.

1. How many roads does B need to catch up? 60 2 = 120 km;

2. How many roads can B catch up per hour? 80-60 = 20 km;

3. How much time does B need to catch up? 120 20 = 6 hours.

My method is arithmetic, the composite formula: 60 2 (80-60) = 6 hours.