At any time t , a particle has a position vector (t^2+1)i+(3t-2)j , relative to the origin O . Show that the acceleration is always directed horizontally . Hence , find the angle between the velocity and acceleration of the particle at any time , t .

v=2ti+3j , a=2i

Since a=2i , the particle is always accelerating along the x-axis .

$\displaystyle \cos \theta=\frac{2t(2)+3(0)}{\sqrt{4t^2+9}\cdot \sqrt{4}}$

$\displaystyle =\frac{2t}{\sqrt{4t^2+9}}$

so am i correct ?