Question

X=-2+4t-t^2 Найти Xo; Vox; Ax Записать ур-ие Vx(t) Написать ур-ие Sx(t) Найти координату тела через 2 с после начала движения

Answer 1

Х₀=-2 м, V₀x=4 м/с, ax=-2 м/с²
Vx(t)=4-2t
Sx(t) =4t-t²
x(2)=-2+4*2 -2² =2м

Answer 2

I can help you with your math problem. Here is the solution:

- Xo is the initial position of the body. It is equal to the value of X when t = 0. Therefore, Xo = -2. - Vox is the initial velocity of the body. It is equal to the derivative of X with respect to t at t = 0. Therefore, Vox = 4 - 2t|_{t=0} = 4. - Ax is the acceleration of the body. It is equal to the second derivative of X with respect to t. Therefore, Ax = -2. - Vx(t) is the equation of velocity as a function of time. It is obtained by integrating Ax with respect to t and using the initial condition Vox. Therefore, Vx(t) = -2t + 4. - Sx(t) is the equation of displacement as a function of time. It is obtained by integrating Vx(t) with respect to t and using the initial condition Xo. Therefore, Sx(t) = -t^2 + 4t - 2. - The coordinate of the body after 2 seconds of motion is equal to the value of X when t = 2. Therefore, X(2) = -4 + 8 - 2 = 2.Ответь+подробно.+X=-2+4t-t^2+Найти+Xo;+Vox;+Ax+Записать+ур-ие+Vx(t)+Написать+ур-ие+Sx(t)+Найти+координату+тела+через+2+с+после+начала+движения