Question

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

• A. square of diplacement
• 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),

$-a=v^3\left(2a\right)$

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