# Chapter 3 - Motion in Two or Three Dimensions - Problems - Exercises - Page 96: 3.41

When $t = \frac{2b}{3c}$, the velocity vector makes an angle of $45.0^{\circ}$.

#### Work Step by Step

$x(t) = bt^2$ $v_x(t) = \frac{dx}{dt} = 2bt$ $y(t) = ct^3$ $v_y(t) = \frac{dy}{dt} = 3ct^2$ If the velocity vector makes an angle of $45.0^{\circ}$, then: $v_x(t) = v_y(t)$ $2bt = 3ct^2$ $t = \frac{2b}{3c}$. When $t = \frac{2b}{3c}$, the velocity vector makes an angle of $45.0^{\circ}$.

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