Question

The diplacement '×' of a particle (in metre) is related to time(in second) by the relation x=2t^3-3t^2+2t+2. Then velocity of the particle at the end of 2 second is?​

Answer 1

Answer:

The required velocity is 14 m/s

Explanation:

Given, x(t) = 2t³ - 3t² + 2t + 2

Velocity is the rate of change of displacement of an object.  

We get the velocity by differentiating the position of the object.

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

 

Differentiating,

$\sf v=\dfrac{d}{dt}(2t^3-3t^2+2t+2) \\\\ \sf v=\dfrac{d}{dt}(2t^3)-\dfrac{d}{dt}(3t^2)+\dfrac{d}{dt}(2t)+\dfrac{d}{dt}(2) \\\\ \sf v=6t^2-6t+2+0 \\\\ \sf v=6t^2-6t+2$

Put t = 2 s, to find the velocity at the end of 2 seconds

 v = 6(2)² - 6(2) + 2

 v = 6(4) - 12 + 2

 v = 24 - 10

 v = 14 m/s

∴ The required velocity is 14 m/s