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.

Answer 1

Hint: First derivative of function ‘x’ with respect to time gives velocity and double derivative of ‘x’ gives acceleration.

Complete step-by-step answer:
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.