Question

The position of a particle is given by $r=3.0t\hat{i}-2.0t^2\hat{j}+4.0\hat{k}m$ where t is in seconds and the coefficients have the proper units for r to be in metres.

A) Find the v and a of the particle ?

B) What is the magnitude and direction of velocity of the particle at t = 2.0 s ?

Answer 1

A) Find the derivative of position with time and the derivative of velocity with time.
Given, position of the particle
$\overrightarrow{r}=3.0\hat{i}-2.0t2\hat{j}+4.0\hat{k}m$
As velocity $\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}$,
Therefore, $\overrightarrow{v}=\frac{d}{dt}(3.0t\hat{i}-2.0t^2\hat{j}+4.0\hat{k})$
$\overrightarrow{v}=3.0\hat{i}-4.0t\hat{j}....\left(1\right)$

And acceleration a of the particle will be,
$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}$
Substituting the value of $\overrightarrow{v}$ from equation (1), we get
$\overrightarrow{a}=\frac{d}{dt}(3.0\hat{i}-4.0t\hat{j})$
$\overrightarrow{a}=-4.0\hat{j}...\left(2\right)$

B) Find the value of v at given time and find tan $\theta$.
Putting t=2 in equation (1), we get
$\overrightarrow{v}=3.0\hat{i}-8.0\hat{j}$
The magnitude of velocity, $|\overrightarrow{v}|=\sqrt{3^2+(-8{)}^2}$
$|\overrightarrow{v}|=\sqrt{9+64}$
$|\overrightarrow{v}|=8.54{\text{ms}}^{-1}$
Direction of velocity is given by the angle $\left(\theta \right)$ which it makes with x - axis.
$\tan \theta =-\frac{8}{3}$
or, $\theta ={\tan }^{-1}(-\frac{8}{3})=-{69.44}^{\circ }\approx -{70}^{\circ }$
The negative sign indicates that the direction of velocity is below the x-axis.