The equation of motion of the point is: x = 3T, y = 4t-5t & # 178; find the tangential acceleration when t = 0

The equation of motion of the point is: x = 3T, y = 4t-5t & # 178; find the tangential acceleration when t = 0

vx=x'=3
vy=y'=4-10t
v=√(x'²+y'²)=√(9+(4-10t)^2)
Tangential acceleration
at=v'=(1/2)2(4-10t)(-10)/√(9+(4-10t)^2)
at0=-8

(t+3t²-3)-(-t+4t²）

(t+3t²-3)-(-t+4t²）
=t+3t²-3+t-4t²
=-t²+2t-3

Decomposition factor: (1-7t-7t2-3t3) (1-2t-2t2-t3) - (T + 1) 6=______ ．

Let (T + 1) 3 = x, y = T2 + T + 2, then the original formula = [2 (T2 + T + 2) - 3 (1 + 3T + 3t2 + T3)] [(T2 + T + 2) - (1 + 3T + 3t2 + T3)] - [(T + 1) 3] 2 = (2y-3x) (Y-X) - x2 = 2x2-5xy + 2Y2 = (2x-y) (x-2y) = [2 (T3 + 3t2 + 3T + 1) - (T2 + T + 2)] [(T3 + 3t2 + 3T + 1) - 2 (

The particle moves along the x-axis, v = 1 + 3T & # 178; (SI). T = 0, the particle is at the origin. Find the acceleration a, the motion equation of the particle

1、a=dv/dt=6t(m/s^2)；
2、s=Svdt=S(1+3t^2)dt=(t+t^3)+C,
When t = 0, s = 0, C = 0,
So: S = t + T ^ 3 (m)