Question

If displacement of a particle is given by x= t3 + t² . Then find out velocity and acceleration at time t = 2.​

Answer 1

Answer:

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Given :

Displacement of particle is given by

$\large\bold\red{\sf\:x(t) = {t}^{3} + t^2}$

To Find :

Velocity and acceleration at t = 2sec

Theory :

• Velocity

The rate of change of displacement of a particle with time is called velocity of the particle.

$\large\bold\red{\rm\:Velocity=\dfrac{Distance}{Time\:interval}}$

$\large\bold\red{Velocity = \dfrac{dx}{dt} }$

• Accelaration

It is the rate of change of Velocity with time

$\large\bold\red{\rm\:Acceleration=\dfrac{Velocity}{Time\:interval}}$

$\large\bold\red{\rm\:Acceleration= \dfrac{dv}{dt} }$

Solution :

Displacement of particle given by

$\large\bold\red{\sf\:x(t) = {t}^{3} + t^2}$

Part -1

We have to find the velocity of the particle at t =2 sec

$\large\bold\red{\sf\:x(t) = {t}^{3} + t^2}$

Now differnatiate with respect to t

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

We know that

$\large\bold\red{\sf\dfrac{d(x^n)}{dx}=nx^{n-1}}$

Then ,

$\large\bold\red{\sf\:v= {3t}^{2} + 2t}$

If t = 2 sec

Then,

$\sf\:v=3 \times {2}^{2} + 2 \times 2$

$\sf\:v=12 + 4$

$\large\bold\red{\sf\:v=16 \: ms^{-1}}$

Hence ,The Velocity of a particle At t =2 sec is 16 m/s.

Part -2

We have to find the accelaration of the particle at t = 2 sec

We have ,

$\large\bold\red{\sf\:v= {3t}^{2} + 2t}$

Now , Differentiate it with respect to t

$\sf\implies\dfrac{dv}{dt}= \: \dfrac{d}{dt} {3t}^{2} + \dfrac{d}{dt} 2t$

$\sf\implies\dfrac{dv}{dt}= \: 6t \: + 2$

If t = 2 sec, then accelaration

$\sf\implies\:a=14 \: ms^{-2}$

Hence , the accelaration of the particle Is 14 m/s²

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More information about topic

Both accelaration and Velocity are vector quantities.

The velocity of an object can be positive, zero and negative.

SI unit of Velocity is m/s

SI unit of accelaration is m/s²