The retardation fo a moving particle if the relation between time and position is ` t= A x^3 + Bx^2 ` where ` A` and ` B` are appropriate constants will be .

Options

A. (a)` (6 A x + 2 B)/((3 A x^2 + 2 B x)^3)`
B. (b) ` (6 B x + 6 A)`/((3 Ax^2 + 2 B x)^3)`
C. (c ) `( 6 A+ 2 B x)/((3 A x+2 B x^2)^3)`
D. (d) (6 A+2 B x)/( (3 A+ 2 B x^3)^2)`

Answer

A

Explanation

Here, ` t= A x^3 + B x^2 , (dt)/(dx) (Ax^3 + B x^2)`
` (dt)/(dx) = 3 A x^2 + 2 B x`,
` v = (dx)/(dt) = ( Ax^2 + 2 B x)^1`
Now ` (dv)/(dx) = (-1) [3 A x^2 + 2 B x ]^(-2)` xx (6 A + 2 B0`
Acceleration ` a = (dv)/(dt) = (dv)/(dx) xx (dx)/(dt) = v (dv)/(dx)`
` a = 1/(3 A x^2+2 B) xx ((- 10 ( 6 A x + 2 B))/((3 A x^2 +2 B x)^2 )`
` =(- (6 A x + 2 B))/((3 A x^2 + 2 B x)^3)`
Retardation `=- a= (6 a x + 2 B)/((3 A x^2 + 2 B x )^3)`.