# 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.

Last updated date: 26th Mar 2023

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Answer

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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.

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.

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