Question

If, $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{r}}\mathbf{=}\mathit{b}{\mathit{t}}^{\mathbf{2}}\hat{\mathbf{i}}\mathbf{+}\mathit{c}{\mathit{t}}^{\mathbf{3}}\hat{\mathbf{j}}$ where b and c are positive constants, when does the velocity vector make an angle of 45.0° with the x- and y-axes?

Short answer

The time at which velocity makes 45° angle is $\frac{2b}{3c}$.

Step-by-step solution

Identification of given data

The given data can be listed below,

• The position vector is,$\stackrel{\rightharpoonup }{\mathrm{r}}=b{t}^2\hat{\mathrm{i}}+c{t}^3\hat{\mathrm{j}}$
• The angle made with x and y axis is,$\theta ={45}^o$

Concept/Significance of position vector

The placement of one item with relation to another is determined by its position vector. Position vectors typically begin at the origin and end at any other random point.

Determination of time at which the velocity vector make an angle of 45.0° with the x- and y-axes

The velocity vector is given by,

$\stackrel{\rightharpoonup }{v}=\frac{d\stackrel{\rightharpoonup }{r}}{dt}$

Here,$\stackrel{\rightharpoonup }{r}$ is the position vector.

Substitute values in the above,

$\begin{array}{rcl}\stackrel{\rightharpoonup }{v}& =& \frac{d\left(d{t}^2\hat{i}+c{t}^3\hat{j}\right)}{dt}\\ & =& \left(2bt\hat{i}+3c{t}^2\hat{j}\right)\end{array}$

The angle of velocity is given by,

$\tan \theta =\frac{v_y}{v_x}$

Substitute all the values in the above the time is given by,

$\begin{array}{rcl}\tan {45}^o& =& \frac{3c{t}^2}{2bt}\\ 1& =& \frac{3ct}{2bt}\\ t& =& \frac{2b}{3c}\end{array}$

Thus, the time at which velocity makes 45° angle is $\frac{2b}{3c}$.