Question

if a position x=6t^2-t^3,then find out position of particle when velocity is maximum?​

Answer 1

32

EXPLANATION IS IN THE ATTACHMENT

CONCEPT USED IS MAXIMA AND MINIMA

MARK MY ANSWER AS BRAINLIEST.

Answer 2

Explanation:

x= 6t^2 - t^3 = t^2( 6 -- t) ........................( 1)

Differentiating with respect to t we get,

dx/dt = 12t -3t^2

v = dx/dt is maximum when dx/dt = 0 this leads to

t = 4 sec.

Again d^2x/dt^2 = 12 -- 6t is less than 0 at t = 4,

Therefore v is maximum at t = 4

Then position x = 16× 2 from (1)

= 32 m.

Ans 32 meter.