Question

Тело движется по криволинейной траектории, имеющей радиус кривизны R = 10 м. Закон движения тела S = Вt + Сt2, где С = 2 м/с2, B = 4 м/с. Определить линейную v и угловую ω скорость, а также нормальное ускорение a тела в момент времени t = 3 c.

Answer 1

Первая производная пути (S) равна скорости (V)

V = (S)' = (Bt + Ct^2)' = B + Ct

Угловая скорость = V/R

Нормальное ускорение = V^2/R

Подставь B, C, время и радиус

Answer 2

I can help you with this question. The body moves along a curved trajectory with a radius of curvature R = 10 m. The law of motion of the body is S = Bt + Ct2, where C = 2 m/s2, B = 4 m/s. To determine the linear v and angular ω velocity, as well as the normal acceleration a of the body at the time t = 3 s, we need to use the following formulas:

- v = dS/dt = B + 2Ct - ω = v/R = (B + 2Ct)/R - a = v2/R = (B + 2Ct)2/R

Substituting the given values, we get:

- v = 4 + 2*2*3 = 16 m/s - ω = 16/10 = 1.6 rad/s - a = 162/10 = 25.6 m/s2

Therefore, the linear velocity of the body is 16 m/s, the angular velocity is 1.6 rad/s, and the normal acceleration is 25.6 m/s2 at the time t = 3 s. I hope this helps.