Question

The position of a particle is given by $r=\text{3}\text{.0t}\widehat{i}+2.\text{0}{{\text{t}}^{2}}\widehat{j}+5.0\widehat{k}$ where ‘t’ is in seconds and the coefficients have their proper units for ‘r’ to be in meters. Find u(t) and a(t) of the particle.

Answer 1

Hint: In the question the position of the particle is given which is a function of time. We are asked to determine the speed of the particle as a function of time as well as the acceleration of the particle. Hence differentiating the position vector with respect to time will yield us the answer.
Formula used:
$u(t)=\dfrac{dr}{dt}$
$a(t)=\dfrac{{{d}^{2}}r}{d{{t}^{2}}}$

Complete answer:
Let us say we have a particle whose position vector ‘r’ is given with respect to time. Hence by differentiating the position vector with respect to time the velocity vector u(t) of the particle is obtained as,
$u(t)=\dfrac{dr}{dt}....(1)$
The acceleration a(t) of the vector is nothing but the derivative of the velocity vector. Hence the acceleration of the particle with respect to time is given by,
$\begin{align}
  & a(t)=\dfrac{d}{dt}\left( u(t) \right) \\
 & \Rightarrow a(t)=\dfrac{d}{dt}\left( \dfrac{dr}{dt} \right) \\
 & \therefore a(t)=\dfrac{{{d}^{2}}r}{d{{t}^{2}}}....(2) \\
\end{align}$
Therefore from equation 1 we get the velocity of the particle with respect to time as,
$\begin{align}
  & u(t)=\dfrac{dr}{dt} \\
 & \Rightarrow u(t)=\dfrac{d(\text{3}\text{.0t}\widehat{i}+2.\text{0}{{\text{t}}^{2}}\widehat{j}+5.0\widehat{k})}{dt} \\
 & \therefore u(t)=(\text{3}\text{.0}\widehat{i}+4.\text{0t}\widehat{j})m{{s}^{-1}} \\
\end{align}$
Similarly the acceleration of the particle from equation 2 is given by,
$\begin{align}
  & a(t)=\dfrac{{{d}^{2}}r}{dt} \\
 & \Rightarrow a(t)=\dfrac{{{d}^{2}}(\text{3}\text{.0t}\widehat{i}+2.\text{0}{{\text{t}}^{2}}\widehat{j}+5.0\widehat{k})}{{{d}^{2}}t} \\
 & \Rightarrow a(t)=\dfrac{d\left( \text{3}\text{.0}\widehat{i}+4.\text{0t}\widehat{j} \right)}{dt} \\
 & \therefore a(t)=4.\text{0}\widehat{j}\text{ }m{{s}^{-2}} \\
\end{align}$
Therefore the speed of the particle is equal to $(\text{3}\text{.0}\widehat{i}+4.\text{0t}\widehat{j})m{{s}^{-1}}$ and the acceleration of the particle is equal to $4.\text{0}\widehat{j}\text{ }m{{s}^{-2}}$ .

Note:
From the above results we can conclude that the particle does change its position with respect to the z-axis. The velocity of the particle is constant in the x direction whereas it changes linearly with y direction. From this itself we can conclude that the particle experiences acceleration in the y direction. The equation for acceleration also proves that there is an acceleration in y direction only. It is also to be noted that the velocity and acceleration are in SI units as the position vector and time is also in SI units.