Question

Влосипедист сспускается с горы с постоянным ускорением 1, 2м/с. Какова скорость его движения через 2 мин если начальная скорость была 3м/с?

Answer 1

Найдем конечную скорость V=V0+a*t=3+1,2*120=147 м/с

Только НЕДОУМОК мог придумать такую задачу, потому что это очень большая скорость
Скорее всего в задаче ускорениме а=0,12 м/с2

Answer 2

Problem Analysis

A cyclist is descending a hill with a constant acceleration of 1.2 m/s^2. We need to find the cyclist's velocity after 2 minutes, given that the initial velocity was 3 m/s.

Solution

To solve this problem, we can use the kinematic equation that relates velocity, acceleration, and time:

Final velocity (v) = Initial velocity (u) + (acceleration (a) * time (t))

Let's substitute the given values into the equation:

v = u + (a * t)

Given: - Initial velocity (u) = 3 m/s - Acceleration (a) = 1.2 m/s^2 - Time (t) = 2 minutes = 2 * 60 seconds = 120 seconds

Substituting the values into the equation:

v = 3 m/s + (1.2 m/s^2 * 120 s)

Simplifying the equation:

v = 3 m/s + 144 m/s

Calculating the final velocity:

v = 147 m/s

Therefore, the cyclist's velocity after 2 minutes is 147 m/s.

Answer

The cyclist's velocity after 2 minutes, given an initial velocity of 3 m/s and a constant acceleration of 1.2 m/s^2, is 147 m/s.