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# 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. Verified
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Hint: First derivative of function ‘x’ with respect to time gives velocity and double derivative of ‘x’ 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.
Last updated date: 25th Sep 2023
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