Question

Жук ползет вверх по стволу дереву скорость 6см/с. По тому же дереву ползет вниз гусеница. Сейчас она находиться на 60см ниже жука. С какой скоростью ползет гусеница ,если через 5с растояние между ними будет 100см?решите НЕ УРОВНЕНИЕМ.

Answer 1

1)6*5=30(см)-прополз жук за 5 с
2)100-30=70(см) проползла гусеница за 5 с
3)70/5=14(см) скорость гусеницы

Answer 2

Problem Analysis

We have a beetle crawling up a tree trunk at a speed of 6 cm/s, and a caterpillar crawling down the same tree trunk. The caterpillar is currently 60 cm below the beetle. We need to find the speed at which the caterpillar is crawling if the distance between them becomes 100 cm after 5 seconds.

Solution

To solve this problem, we can use the concept of relative motion. Let's assume that the speed of the caterpillar is x cm/s. Since the beetle is crawling up and the caterpillar is crawling down, their relative speed will be the sum of their individual speeds. Therefore, the relative speed between the beetle and the caterpillar is 6 cm/s + x cm/s.

We know that the distance between them is currently 60 cm, and we need to find the speed at which the distance becomes 100 cm after 5 seconds. We can use the formula distance = speed × time to calculate the distance traveled by each of them.

For the beetle: - Distance traveled by the beetle = 6 cm/s × 5 s = 30 cm

For the caterpillar: - Distance traveled by the caterpillar = (6 cm/s + x cm/s) × 5 s = 30 cm + 5x cm

According to the problem, the distance between them will be 100 cm after 5 seconds. Therefore, we can set up the equation:

30 cm + 5x cm = 100 cm

Now, let's solve this equation to find the value of x.

Calculation

30 cm + 5x cm = 100 cm

Subtracting 30 cm from both sides: 5x cm = 70 cm

Dividing both sides by 5 cm: x cm = 14 cm/s

Answer

Therefore, the caterpillar is crawling at a speed of 14 cm/s.

Verification

Let's verify our answer using the given information. After 5 seconds, the beetle will have crawled up by 30 cm (6 cm/s × 5 s), and the caterpillar will have crawled down by 70 cm (14 cm/s × 5 s). The difference between their positions will be 100 cm, as required.

This solution is based on the assumption that the beetle and the caterpillar are crawling at a constant speed throughout the given time period.