Question

A particle traveling on a road according to equation

X=5T³-2T²+3

and find out

(1) velocity of particle at 3 sec. (2) acceleration of particle at 3 sec.​

Answer 1

velocity = dx/dt

           = d(5t³ - 2t² + 3)/dt

           = 15t² - 4t

At t = 3,   velocity = 15(3)² - 4(3)

                             = 135 - 12

                             = 123 m/s

Acceleration = dv/dt

                   = d(15t² - 4t)/dt

                   = 30t - 4

At t = 3,   a  = 30(3) - 4 = 86 m/s²

Answer 2

Explanation:

dx/dt = 15t² - 4t = velocity

at t = 3, velocity = 15(3)²-4(3)=123 m/s

a = dv/dt = 30t - 4

At t=3, a = 30(3)-4 = 86m/s²

1st answerer has mistakenly put it 2*, correct answer is 86 m/s²

Check it out