Question

The position $\overrightarrow{r}=(2.00t^3-5.00t)\hat{i}+(6.00-7.00t^4)\hat{j},with\overrightarrow{r}$ in meters and t in seconds.
What is the angle between the positive direction of the x-axis and a line tangent to the particle's path at t=2.00 s?

• A. $ta{n}^{-1}\left(\frac{28}{6}\right)$
• B. $ta{n}^{-1}\left(\frac{-224}{19}\right)$
• C. $ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• D. ${45}^{\circ }$

Answer 1

The correct option is B

$ta{n}^{-1}\left(\frac{-224}{19}\right)$

$\overrightarrow{r}=(2t^3-5t)\hat{i}+(6-7t^4)\hat{j}$
$\overrightarrow{v}=\frac{dr}{dt}=(6t^2-5)\hat{i}+(-28t^3)\hat{j}$
At t = 2
$\overrightarrow{v}=19\hat{i}+(-224)\hat{j}$
Now the direction of v is the direction of tangent.

Whatever the path be, the instantaneous velocity is always tangential to it.
so that angle between velocity and x-axis be the angle between tangent and x-axis.
The angle is the slope of velocity vector
$\Rightarrow tan\theta =\frac{v_y}{v_x}=\frac{-224}{19}$
$\Rightarrow \theta =ta{n}^{-1}\left(\frac{-224}{19}\right)$