已知质点的运动学方程为:r=(5+2t-0.5t(平方))i+(4t+t(三次方)/3)j(SI)(1).当t=2s时,(2).加速度与x轴正方向的夹角?

Question

已知质点的运动学方程为:r=(5+2t-0.5t(平方))i+(4t+t(三次方)/3)j (SI) (1).当t=2s时,(2).加速度与x轴正方向的夹角?

Answer 1

:r=(5+2t-0.5t(平方))i+(4t+t(三次方)/3)j (SI)
x=(5+2t-0.5t^2),y=(4t+t^3/3)
Vx=dx/dt=2-t Vy=dy/dt=4+4t^2/3
ax=dVx/dt=-1,ay=dVy/dt=4t/3
t=2S ax=-1,ay=4t/3=8/3
a^2=ax^2+ay^2=1+64/9=73/9
a=(1/3)*根号（73）