Question

If a particle moving along a line following the law $t=a{s}^2+bs+c$ then the retardation of the particle is proportional to

• A. Square of displacement
• B. Square of velocity
• C. Cube of displacement
• D. Cube of velocity

Answer 1

The correct option is D Cube of velocity

Consider,

$t=a{s}^2+bs+c$

Differentiate w.r.t. s

$\frac{dt}{ds}=2as+b$

Or,

$v=\frac{ds}{dt}=(2as+b{)}^{-1}$ ------- (i)

$Retardation=-a=\frac{-dv}{dt}$

Or, $-a=\frac{dv}{ds}\times \frac{ds}{dt}$

$-a=v\times \frac{dv}{ds}$

$-a=v\times \frac{d}{ds}(2as+b{)}^{-1}$

$-a=v(2as+b{)}^{-2}(2a)$

Putting the value of $v$ from (i),

$-a=v(2as+b{)}^{-3}(2a)$

From this, it is evident that the retardation of the particle is proportional to the cube of the velocity. Option D.