Question

find the velocity and acceleration if the position of particle is given by X=6t² + 5t³ for t=2s?​

Answer 1

$\huge{\mathbb{\red{ANSWER:-}}}$

Given :-

$\sf{position \: of \: a \: particle -}$

$\sf{x = 6t^{2} + 5t^{3}}$

To Find :-

$\sf{Velocity \: and \: acceleration \: of \: the \: particle = ?}$

Using Formulas :-

$\sf{Velocity (V) =\dfrac{dx}{dt}}$

$\sf{Acceleration (a) =\dfrac{dV}{dt}}$

Explanation :-

$\sf{x = 5t^{3} + 6t^{2}}$

Differential with respect to t -

$\sf{\dfrac{dx}{dt} = 5(3t^{2}) + 6(2t)}$

$\sf{\dfrac{dx}{dt} = 15t^{2} + 12t}$

$\sf{V = 15t^{2} + 12t}$

at , t = 2 Second

$\sf{V = 15(4) + 12(2)}$

$\sf{V = 60 + 24}$

$\sf{V = 84 \: m/Second}$

Now ,

$\sf{V = 15t^{2} + 12t}$

Differential with respect to t -

$\sf{\dfrac{dV}{dt} = 15(2t) + 12}$

$\sf{a = 30t + 12}$

at , t = 2 Second

$\sf{a = 30(2) + 12}$

$\sf{a = 60 + 12}$

$\sf{a = 72 \: m/Sec^{2}}$

As We got :-

$\sf{Velocity (V) = 84 \: m/Second}$

$\sf{Acceleration (a) = 72 \: m/Sec^{2}}$

Answer 2

Question :

find the velocity and acceleration if the position of particle is given by x=6t² + 5t³ for t=2s?

Answer :

$\text{Velocity = 84 m/s} \\ \text{Acceleration = 72 } m/s^2$

Explanation :

        the position of the particle is given by,

                   x = 6t² + 5t³

we have to find the velocity and acceleration of the particle at t = 2 s

=> differentiate position to get velocity

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

             

              $v=\frac{d}{dt}(6t^2+5t^3) \\\\ v=\frac{d}{dt}(6t^2)+\frac{d}{dt}(5t^3) \\\\ {\bf v=12t+15t^2} \\\\ v=12(2)+15(2^2)\\\\v=24+15(4)\\\\v=24+60\\\\v=84 \ m/s$

∴ Velocity = 84 m/s

=> differentiate velocity to get acceleration.

                   $\boxed{\bf a=\frac{dv}{dt}}$

                  $a=\frac{d}{dt}(12t+15t^2) \\\\ a=\frac{d}{dt}(12t)+\frac{d}{dt}(15t^2) \\\\ a=12+30t \\\\ a=12+30(2) \\\\ a=12+60\\\\a=72 \ m/s^2$

∴ Acceleration = 72 m/s²