Question

Тіло рухається прямолінійною ділянкою шляху, а його швидкість (у м/с) при цьому змінюється за законом v(t)= 6t²+2t-2. На якій відстані (у м) від початку відліку знаходилося тіло через 1 с після початку руху, якщо через 2 с після початку руху ця відстань становила 18 м?​

Answer 1

Ответ:

через 1 с після початку руху тіло знаходилось на відстані 3 м від початку відліку.

Объяснение:

Знайдемо спочатку значення сталої інтегрування C, використавши відомі значення шляху:

x(2) = 2(2)³+2²(2)-2(2) + C = 18 м

C = 18 - 16 - 4 + 8 = 6 м

Тепер, використовуючи це значення C, можемо знайти відстань x через 1 с після початку руху:

x(1) = 2(1)³+1²(1)-2(1) + 6 = 3 м

Отже, через 1 с після початку руху тіло знаходилось на відстані 3 м від початку відліку.

Answer 2

Problem Analysis

We are given that the body is moving along a straight path, and its velocity is changing according to the equation v(t) = 6t² + 2t - 2. We need to find the distance traveled by the body from the starting point after 1 second and after 2 seconds.

Solution

To find the distance traveled by the body, we need to integrate the velocity function with respect to time. The integral of velocity gives us the displacement or distance traveled.

Let's integrate the velocity function v(t) = 6t² + 2t - 2 with respect to time t to find the displacement function s(t):

∫v(t) dt = ∫(6t² + 2t - 2) dt

To integrate each term, we use the power rule of integration:

∫t^n dt = (t^(n+1))/(n+1) + C

where C is the constant of integration.

Integrating the first term 6t²: ∫6t² dt = (6/3)t^3 + C1 = 2t^3 + C1

Integrating the second term 2t: ∫2t dt = (2/2)t^2 + C2 = t^2 + C2

Integrating the third term -2: ∫-2 dt = -2t + C3

Now, let's combine the results: s(t) = 2t^3 + t^2 - 2t + C

To find the constant of integration C, we need to use the initial condition. We know that at t = 0, the distance traveled is 0 (since it's the starting point). So, we can substitute t = 0 and s(t) = 0 into the displacement function:

0 = 2(0)^3 + (0)^2 - 2(0) + C 0 = 0 + 0 - 0 + C C = 0

Therefore, the displacement function becomes: s(t) = 2t^3 + t^2 - 2t

Now, we can find the distance traveled by the body after 1 second and after 2 seconds.

1. After 1 second: To find the distance traveled after 1 second, we substitute t = 1 into the displacement function s(t): s(1) = 2(1)^3 + (1)^2 - 2(1) s(1) = 2 + 1 - 2 s(1) = 1 meter

Therefore, the body is located 1 meter from the starting point after 1 second.

2. After 2 seconds: To find the distance traveled after 2 seconds, we substitute t = 2 into the displacement function s(t): s(2) = 2(2)^3 + (2)^2 - 2(2) s(2) = 16 + 4 - 4 s(2) = 16 meters

Therefore, the body is located 16 meters from the starting point after 2 seconds.

Answer

The body is located 1 meter from the starting point after 1 second and 16 meters from the starting point after 2 seconds.