Path of a particle moving in x-y plane is y = 3x + 4. At some instant suppose x-component of velocity is 1 m/s and it is increasing at a constant rate of 1 m/s². Then, at this instant

$$\begin{gathered} y=3x+4 \\ Differentiating wrt time \\ {v}_y=3v_x=3\left(1\right)=3 m{s}^{-1} \\ Differentiating wrt time \\ {a}_y=3a_x=3\left(1\right)=3 m{s}^{-2} \\ now, v_{net}=\sqrt{v_x^2+v_y^2}=\sqrt{10}m{s}^{-1} \\ {a}_{net}=\sqrt{a_x^2+a_y^2}=\sqrt{10}m{s}^{-2} \end{gathered}$$

Acceleration is constant . So v-t curve will be straight line.