Question

Движение тела задано уравнением x=t+0.05t^2 .Определить ax,V0x, скорость тела через t=10c. Построить графики Vx(t), ax(t), примерный Sx(t). Каково направление векторов ускорения и скорости?

Answer 1

Если записать уравнение равноускоренного движения в общем виде, то получим: x(t) = Xo + Vo*t + a/2*t²

ДАНО:  x(t) = t + 0.05*t²

РЕШЕНИЕ.

Просто глазами из формулы:

Хо = 0 м - начальное смещение.  

V0 = 1 м/с - начальная скорость,

а = 2*0,05 = 0,1 м/с² - ускорение.

Скорость по формуле:  V(t) = Vo + a*t.

Вычисляем при t = 10 c.   V(10) = 1 + 0.1*10 = 2 м/с - скорость - ОТВЕТ

Значения и скорости и ускорения положительные, значит и их вектора в положительном направлении.

Графики функций на рисунке в приложении.

Answer 2

Equation of Motion and Determining ax and V0x

The equation of motion for the given body is given by x = t + 0.05t^2. To determine the acceleration ax and initial velocity V0x, we need to differentiate the equation of motion with respect to time.

Differentiating x = t + 0.05t^2 with respect to time, we get:

dx/dt = 1 + 0.1t

This represents the velocity Vx of the body as a function of time t. To find the acceleration ax, we differentiate the velocity equation with respect to time:

d^2x/dt^2 = 0.1

Therefore, the acceleration ax is constant and equal to 0.1.

To find the initial velocity V0x, we substitute t = 0 into the velocity equation:

V0x = 1 + 0.1(0) = 1

So, the initial velocity V0x is 1.

Graphs of Vx(t), ax(t), and Sx(t)

To plot the graphs of Vx(t), ax(t), and Sx(t), we can use the equations derived above.

The graph of Vx(t) represents the velocity of the body as a function of time. Since the velocity Vx is given by dx/dt = 1 + 0.1t, the graph will be a straight line with a slope of 0.1 and a y-intercept of 1.

The graph of ax(t) represents the acceleration of the body as a function of time. Since the acceleration ax is constant and equal to 0.1, the graph will be a horizontal line at a height of 0.1.

The graph of Sx(t) represents the displacement of the body as a function of time. To find an approximate equation for Sx(t), we can integrate the velocity equation:

Sx(t) = ∫(1 + 0.1t) dt = t + 0.05t^2 + C

where C is the constant of integration. Since the equation of motion already includes the constant term, the constant of integration can be ignored. Therefore, the approximate equation for Sx(t) is Sx(t) = t + 0.05t^2.

Direction of Acceleration and Velocity Vectors

The direction of the acceleration vector ax is determined by its sign. In this case, since ax is positive (0.1), the acceleration vector ax is directed in the positive x-direction.

The direction of the velocity vector Vx is determined by its sign as well. Since V0x is positive (1), the velocity vector Vx is also directed in the positive x-direction.

Therefore, both the acceleration and velocity vectors are directed in the positive x-direction.

I hope this helps! Let me know if you have any further questions.