Question

ПОМОГИТЕ ПОЖАЛУЙСТА Сбегая по эскалатору со скоростью v1, мальчик насчитал N1, ступенек. Когда он увеличил скорость на Δv, то насчи- тал в полтора раза больше ступенек. Сколько ступенек N2 насчитает мальчик, спускаясь по неподвижному эскалатору? ​

Answer 1

Ответ:

Дано:

v2=1,5 v1

N- ?

t=hN/v1

t=hN1/v1+v0

hN/v1=hN1/v1+v0

N/v1=N1/v1+v0

N/1,5v=N1+дельтаN/1,5v+v0

v1+v0=v1N1/N

1,5v1+v0=1,5v1N1+дельтаN/N

1,5v1+v0-v1-v0=1,5N1+дельтаN/N-v1N1/N

0,5=1,5N1+1,5дельтаN1-N1/N

N=0,5N1+1,5дельтаN/0,5

N=N1+3дельтаN

Answer 2

Problem Analysis

To solve this problem, we can use the concept of work done by a force. When the boy increases his speed, the work done by the force of gravity on him also increases. We can use this relationship to find the solution.

Solution

Let's denote the initial number of steps counted by the boy as N1 and the initial speed as v1. When the boy increases his speed to v1 + Δv, he counts 1.5N1 steps.

The work done by the force of gravity is proportional to the distance covered. Therefore, the work done when the boy counts N1 steps at speed v1 is equal to the work done when he counts 1.5N1 steps at speed v1 + Δv.

Using the work-energy principle, we can set up the following equation:

Work done at speed v1 = Work done at speed v1 + Δv

The work done by the force of gravity is given by the formula work = force × distance. Since the force of gravity is constant, the work done is directly proportional to the distance covered.

Let's denote the height of each step as h. Then the work done when counting N1 steps at speed v1 is N1 × h and the work done when counting 1.5N1 steps at speed v1 + Δv is 1.5N1 × h.

Setting up the equation:

N1 × h = 1.5N1 × h

Solving for N1:

N1 = 1.5N1

This implies that the number of steps counted by the boy is the same at both speeds. Therefore, the number of steps N2 the boy will count while descending at the increased speed v1 + Δv on a stationary escalator is N1.

Conclusion

The number of steps N2 the boy will count while descending at the increased speed v1 + Δv on a stationary escalator is the same as the number of steps N1 he counted at the initial speed v1.