If a particle moving along a line follows the law t = as2 + bs + c, then the retardation of the particle is proportional to

A
square of the velocity
B
cube of the displacement
C
cube of the velocity
D
square of the displacement

Solution:

we have,

$\begin{array}{l} t = a s^{2} + b s + c \\ \Rightarrow 1 = (2 a s + b) \frac{d s}{d t} \\ \Rightarrow \frac{d s}{d t} = \frac{1}{2 a s} + b \Rightarrow \frac{d^{2} s}{d t^{2}} = - \frac{2 a}{(2 a s + b)^{2}} \frac{d s}{d t} = - 2 a {(\frac{d s}{d t})}^{3} \end{array}$

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

Solution:

Related content