Question

the displacement of a particle along x axis is given by x=4+8t+14t².obtain it's velocity and acceleration at t=2 s​

Answer 1

Given :

Displacement of particle is given by

$\sf\:x(t)=4+8t+14t^2$

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.

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

• Accelaration

It is the rate of change of Velocity with time

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

Solution :

Displacement of particle given by

$\sf\:x(t)=4+8t+14t^2$

Part -1

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

$\sf\:x=4+8t+14t^2$

Now differnatiate with respect to t

$\sf\dfrac{dx}{dt}=\dfrac{d(4)}{dt}+\dfrac{d(8t)}{dt}+\dfrac{d(14t^2)}{dt}$

We know that

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

Then ,

$\sf\:v=8+28t$

If t = 2 sec

Then,

$\sf\:v=28\times2+8$

$\sf\:v=56+8$

$\sf\:v=64ms^{-1}$

Hence ,The Velocity of a particle At t =2 sec is 64 m/s

Part -2

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

We have ,

$\sf\:v=8+28t$

Now , Differentiate it with respect to t

$\sf\implies\dfrac{dv}{dt}=\dfrac{d(8)}{dt}+\dfrac{d(28t)}{dt}$

$\sf\implies\dfrac{dv}{dt}=28$

$\sf\implies\:a=28ms^{-2}$

Hence , the accelaration of the particle Is 28m/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
• Dimension of Velocity: $\sf\:[M^0LT{}^{-1}]$

Answer 2

Answer:

$\huge\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Answer}}}}}}}} \\ \large\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Your~answer↓}}}}}}}}$

Given :

Displacement of particle is given by

$\large\bold\red{\sf\:x(t)=4+8t+14t^2}$

To Find :

♡ Velocity and acceleration at t = 2sec

Formula used

• Velocity

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

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

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

• Accelaration

It is the rate of change of Velocity with time

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

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

$\huge\bold\red{SOLUTION}$

Displacement of particle given by

$\large\bold\red{\sf\:x(t)=4+8t+14t^2}$

Solution of Part  -1

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

sec

$\large\bold\red{\sf\:x(t)=4+8t+14t^2}$

Now differnatiate with respect to t

$\sf\dfrac{dx}{dt}=\dfrac{d(4)}{dt}+\dfrac{d(8t)}{dt}+\dfrac{d(14t^2)}{dt}$

We know that

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

Then ,

$\sf\:v=8+28t$

If t = 2 sec

Then,

$\sf\:v=28\times2+8 \\ </h3><h3>\sf\:v=56+8 \\ </h3><h3>\sf\:v=64ms^{-1}$

Hence ,The Velocity of a particle At t =2 sec is 64 m/s

Solution of Part -2

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

We have ,

$\large\bold\red{\sf\:x(t)=4+8t+14t^2}$

$\large\bold\red{\sf\:v=8+28t}$

Now , Differentiate it with respect to t

$\sf\implies\dfrac{dv}{dt}=\dfrac{d(8)}{dt}+\dfrac{d(28t)}{dt}$

$\large\bold\red{\sf\implies\dfrac{dv}{dt}=28}$

$\large\bold\red{\sf\implies\:a=28ms^{-2}}$

Hence , the accelaration of the particle Is 28m/s²

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