Question

A particle is moving along the path given by $y=\frac{c}{6t}$ ( where c is a positive constant). The relation between the acceleration (a) and the velocity (v) of the particle at $t=5$ sec. is

• A. $5a=-2v$
• B. $a=\sqrt{v}$
• C. $a=5v$
• D. $a=v$

Answer 1

The correct option is A $5a=-2v$

Given,

$y=\frac{c}{6t}$

Velocity, $v=\frac{dy}{dt}$

$v=\frac{d\left(\frac{c}{6t}\right)}{dt}$

$v=-\frac{c}{6t^2}$. . . . . . . . . . . . .(1)

Acceleration, $a=\frac{dv}{dt}$

$a=\frac{d(-\frac{c}{6t^2})}{dt}$

$a=\frac{2c}{6t^3}$. . . . . . . . . . . (2)

Divide equation (2) by equation (1),

$\frac{a}{v}=\frac{\frac{2c}{6t^3}}{-\frac{c}{6t^2}}$

$\frac{a}{v}=-\frac{2c}{6t^3}\times \frac{6t^2}{c}$

$\frac{a}{v}=-\frac{2}{t}$

At $t=5sec$

$\frac{a}{v}=-\frac{2}{5}$

$5a=-2v$