Question

The position x of a particle varies with time according to the relation $$x=t^3+3t^2+2t$$. 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$$\Rightarrow x=t^3+3t^2+2t-(1)$$
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 $$={\displaystyle \frac{dx}{dt}}$$
$$\begin{array}{rl}& \overrightarrow{v}={\displaystyle \frac{d}{dt}}\left(x\right)={\displaystyle \frac{d}{dt}}(t^3+3t^2+2t)\\ & \Rightarrow \overrightarrow{v}=3t^2+2\left(3t\right)+2\\ & \overrightarrow{v}=3t^2+6t+2\end{array}$$
The unit of velocity is meter per second (m/sec).
$$\therefore \overrightarrow{v}=(3t^2+6t+2)$$m/sec.
To find acceleration, which is the rate of change of velocity.
Acceleration, $$\overrightarrow{a}={\displaystyle \frac{d\overrightarrow{v}}{dt}}$$
$$\begin{array}{rl}& \overrightarrow{a}={\displaystyle \frac{d}{dt}}\left(\overrightarrow{v}\right)={\displaystyle \frac{d}{dt}}(3t^2+6t+2)\\ & \overrightarrow{a}=2\times \left(3t\right)+6=6t+6\end{array}$$
The unit of acceleration is meter per second square $$\left(m/{\sec }^2\right)$$.
$$\therefore$$ $$\overrightarrow{a}=(6t+6)m/{\sec }^2$$
$$\therefore$$Velocity of the function, $$\overrightarrow{v}=(3t^2+6t+2)$$m/sec.
 Acceleration of the function, $$\overrightarrow{a}=(6t+6)m/{\sec }^2$$

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