Question

The position x of a particle varies with time t according to the relation x = t^3+3t^2+2t
Find velocity and acceleration as function of time.
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Answer 1

differentiate it and put time 2 at t and add it u will get the answet

Answer 2

Given :

The position x of a particle varies with time t according to the relation :

$x = t^3+3t^2+2t$

To Find :

Velocity and acceleration as function of time.

Solution :

We know , velocity is given as :

$v=\dfrac{dx}{dt}\\\\v=\dfrac{d(t^3+3t^2+2t)}{dt}\\\\v=3t^2+6t+2$

Now , acceleration is given as :

$a=\dfrac{dv}{dt}\\\\a=\dfrac{d(3t^2+6t+2)}{dt}\\\\a=6t+6$

Therefore , velocity and acceleration as function of time is $3t^2+6t+2$ and

$6t+6$ respectively .

Learn More :

Instantaneous velocity and acceleration

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