Answer: [tex]t=\dfrac{b}{c}[/tex]
Explanation:
Given
The position vector is [tex]\vec{r}=bt^2\hat{i}+ct^3\hat{j}\\[/tex]
So, the angle made by position vector is [tex]45^{\circ}[/tex] at
[tex]\Rightarrow \tan 45^{\circ}=\dfrac{r_y}{r_x}\\\\\Rightarrow \tan 45^{\circ}=\dfrac{ct^3}{bt^2}\\\\\Rightarrow bt^2=ct^3\\\\\Rightarrow t=\dfrac{b}{c}[/tex]
At [tex]t=\dfrac{b}{c}[/tex], position vector makes [tex]45^{\circ}[/tex] with x and y axes
