Question

# The position x of a particle varies with time according to the relation $x={{t}^{3}}+3{{t}^{2}}+2t$. Find velocity and acceleration as functions of time.

Hint: First derivative of function â€˜xâ€™ with respect to time gives velocity and double derivative of â€˜xâ€™ gives acceleration.

The position of the particle is given by variable x, and it varies according to time.
Given the relation$\Rightarrow x={{t}^{3}}+3{{t}^{2}}+2t-(1)$
To find the velocity, which is the rate of change of displacement.
The first derivative of Eqn(1) gives us the velocity and the second derivation will give the acceleration.
$\therefore$Velocity $=\dfrac{dx}{dt}$
\begin{align} & \overrightarrow{v}=\dfrac{d}{dt}\left( x \right)=\dfrac{d}{dt}\left( {{t}^{3}}+3{{t}^{2}}+2t \right) \\ & \Rightarrow \overrightarrow{v}=3{{t}^{2}}+2\left( 3t \right)+2 \\ & \overrightarrow{v}=3{{t}^{2}}+6t+2 \\ \end{align}
The unit of velocity is meter per second (m/sec).
$\therefore \overrightarrow{v}=\left( 3{{t}^{2}}+6t+2 \right)$m/sec.
To find acceleration, which is the rate of change of velocity.
Acceleration, $\overrightarrow{a}=\dfrac{d\overrightarrow{v}}{dt}$
\begin{align} & \overrightarrow{a}=\dfrac{d}{dt}\left( \overrightarrow{v} \right)=\dfrac{d}{dt}\left( 3{{t}^{2}}+6t+2 \right) \\ & \overrightarrow{a}=2\times \left( 3t \right)+6=6t+6 \\ \end{align}
The unit of acceleration is meter per second square $\left( m/{{\sec }^{2}} \right)$.
$\therefore$ $\overrightarrow{a}=\left( 6t+6 \right)m/{{\sec }^{2}}$
$\therefore$Velocity of the function, $\overrightarrow{v}=\left( 3{{t}^{2}}+6t+2 \right)$m/sec.
Acceleration of the function, $\overrightarrow{a}=\left( 6t+6 \right)m/{{\sec }^{2}}$

Note: We know velocity$=\dfrac{Displacement}{time}$and acceleration$=\dfrac{velocity}{time}$, here the velocity is taken as the rate of change of displacement w.r.t the time, so differentiation $\left( \dfrac{dx}{dt} \right)$is done.