Question

The relation between time t and distance x is $t=a{x}^2+bx$ where a and b are constants. The retardation is ( if v is velocity of the particle ):

• A. $2$ bv
• B. $2b{v}^2$
• C. $\frac{2a{v}^2}{b}$
• D. none of these

Answer 1

The correct option is C $\frac{2a{v}^2}{b}$

$t=a{x}^2+bx$

Diffrentiate w.r.t time

$1=2ax\frac{dx}{dt}+b\frac{dx}{dt}$......(1)

Diffrentiating again to get acceleration

$0=2a(\frac{dx}{dt}{)}^2+2ax\frac{d^{2}x}{d{t}^2}+b\frac{d^{2}x}{d{t}^2}$

$a=\frac{d^{2}x}{d{t}^2}$

$a=\frac{-2a{v}^2}{2ax+b}$

putting the value of x from (1)

$a=\frac{-2a{v}^2}{2a\times \left(b/2a\right)+b}$

$a=-\frac{2a{v}^2}{b}$

Retardation $=\frac{2a{v}^2}{b}$