Question

The position of a particle is given by X=4t^3-2t^2+3t , find the velicity and acceleration in 3 seconds?



The position of a particle is given by X=4t^3-2t^2+3t , find the velicity and acceleration in 3 seconds?​

Answer 1

Answer:

refer this.

sorry for late answer

Answer 2

Given:

Position of particle, x= 4t³-2t²+3t

To Find:

Velocity and acceleration of particle after 3 second

Solution:

We know that,

• Velocity of a body 'v' is given by

$\pink{\underline{\boxed{\bf{v=\dfrac{dx}{dt}}}}}$

• Acceleration of a body 'a' is given by

$\purple{\underline{\boxed{\bf{a=\dfrac{dv}{dt}}}}}$

where,

x is the displacement of particle at time t

━━━━━━━━━━━━━━━━━━━━━

Let the velocity of given particle be v and acceleration of given particle be a

So,

$\longrightarrow\rm{v=\dfrac{dx}{dt}}$

$\longrightarrow\rm{v=\dfrac{d(4t^{3}-2t^{2}+3t)}{dt}}$

$\longrightarrow\rm{v=4(3t^{2})-2(2t)+3(1)}$

$\longrightarrow\rm{v=12t^{2}-4t+3}$

At t= 3 sec

$\longrightarrow\rm{v=12(3)^{2}-4(3)+3}$

$\longrightarrow\rm{v=12(9)-12+3}$

$\longrightarrow\rm{v=108-12+3}$

$\longrightarrow\rm\green{v=99\ m/s}$

Also,

$\longrightarrow\rm{a=\dfrac{dv}{dt}}$

$\longrightarrow\rm{a=\dfrac{d(12t^{2}-4t+3)}{dt}}$

$\longrightarrow\rm{a=12(2t)-4(1)+0}$

$\longrightarrow\rm{a=24t-4}$

At t= 3 sec

$\longrightarrow\rm{a=24(3)-4}$

$\longrightarrow\rm{a=72-4}$

$\longrightarrow\rm\green{a=68\ m/s^{2}}$

Hence, the velocity of particle is 99 m/s and acceleration of particle is 68 m/s².