Question

The displacement of a particle is given as x=6t^3+2t.calculate the velocity of the body at t=5 seconds

Answer 1

Given:

Displacement of a particle,x= 6t³+2t

To Find:

The velocity of the body at t=5 sec

Solution:

We know that,

• Velocity of a body 'v' is given by

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

where x is displacement or position of particle

━━━━━━━━━━━━━━━━━━━━━━━━━

Let the velocity of particle be v

So,

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

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

$\longrightarrow\rm{v=6(3t^{2})+2(1)}$

$\longrightarrow\rm{v=18t^{2}+2}$

At t= 5 sec

$\longrightarrow\rm{v=18(5)^{2}+2}$

$\longrightarrow\rm{v=18(25)+2}$

$\longrightarrow\rm{v=450+2}$

$\longrightarrow\rm\green{v=452\ m/s}$

Hence, the velocity of particle at t= 5 sec is 452 m/s.

Answer 2

Answer :-

$\longrightarrow$$x = {6t }^{3} + 2t$

$\implies$$v = \frac{dx}{dt}$

$\longrightarrow$$v = {18t}^{2} + 2$

Substituting t = 5

$\implies$v = 18 × 5² + 2

$\implies$v = 18 × 25 + 2

$\implies$v = 450 + 2

$\implies$v = 452

ADDITIONAL INFORMATION :-

$\frac{d(c)}{dx} = 0$

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

$\frac{d(sinx)}{dx} = cosx$

$\frac{d(cosx)}{dx} = - sinx$

$\frac{d(logx)}{dx} = \frac{1}{x}$

Thanks