Question

let the position of the object be x = 2t^2-6t^+1. find its velocity at t =2 sec and between 2 and 3 sec​

Answer 1

Given :

Position equation of particle is given by

• $\tt{x=2t^2-6t+1}$

To Find :

Velocity of object at t = 2s and between 2 and 3 seconds.

Solution :

In order to find velocity of particle we need to differentiate the given position equation$.$

$:\implies\:\underline{\boxed{\bf{\orange{v=\lim\limits_{t\to 0}\:\dfrac{\Delta x}{\Delta t}=\dfrac{dx}{dt}}}}}$

➙ v = dx/dt

➙ v = d(2t² - 6t + 1) / dt

➙ v = 4t - 6

♦ Velocity at t = 2 s :

➛ v = 4t - 6

➛ v = 4(2) - 6

➛ v = 8 - 6

➛ v = 2 m/s

♦ Velocity at t = 3 s :

➛ v = 4t - 6

➛ v = 4(3) - 6

➛ v = 12 - 6

➛ v = 6 m/s

♦ Average velocity b/w t = 2 and 3 s :

➛ v = v(2) + v(3) / 2

➛ v = (2 + 6) / 2

➛ v = 8/2

➛ v = 4 m/s

Answer 2

Given :

Position equation of particle is given by

$\tt{x=2t^2-6t+1}$

To Find :

Velocity of object at t = 2s and between 2 and 3 seconds.

Solution :

In order to find velocity of particle we need to differentiate the given position equation$.$

$:\implies\:\underline{\boxed{\bf{\orange{v=\lim\limits_{t\to 0}\:\dfrac{\Delta x}{\Delta t}=\dfrac{dx}{dt}}}}}$

➙ v = dx/dt

➙ v = d(2t² - 6t + 1) / dt

➙ v = 4t - 6

♦ Velocity at t = 2 s :

➛ v = 4t - 6

➛ v = 4(2) - 6

➛ v = 8 - 6

➛ v = 2 m/s

♦ Velocity at t = 3 s :

➛ v = 4t - 6

➛ v = 4(3) - 6

➛ v = 12 - 6

➛ v = 6 m/s

♦ Average velocity b/w t = 2 and 3 s :

➛ v = v(2) + v(3) / 2

➛ v = (2 + 6) / 2

➛ v = 8/2

➛ v = 4 m/s